For this entire blog post, we fix a nonnegative integer \(n\). We are interested in the following theorem.
Theorem 1. There is a natural isomorphism between the set of morphisms from a scheme \(X\) to \(\mathbb P^n_{\mathbb Z}\) and the set of isomorphism classes of line bundles \(\mathcal L\) equipped with \(n+1\) global sections \(\{s_0,\ldots,s_{n+1}\}\) which have no common zero.
This theorem confused me for a long time because I did not understand where these sections and line bundles should come from. In my opinion, many classical texts on the subject do not provide satisfactory explanations as to why a line bundle \(\mathcal L\) is necessary (instead of merely working with global sections of \(\mathcal O_X\)). (It is not hard to explain the presence of the \(n+1\) sections: this is just a fancy way of asking for a surjective map \(\mathcal O_X^{n+1}\to\mathcal L\).)
For my own personal use, we will try to explain “what’s going on” behind Theorem 1, meaning that we would like to explain why \[X\mapsto\left\{\mathcal O^{n+1}\twoheadrightarrow\mathcal L:\mathcal L\in\operatorname{Pix}X\right\}\] is a natural thing to do. We will also take a “functor of points” perspective, pretending that we have never seen the construction of the scheme \(\mathbb P^n_{\mathbb Z}\) and instead try to characterize it as representing the above functor; at the end, we will construct it. Thus, we will eventually provide a proof of Theorem 1, but it is far from the most efficient one.
Intuitively, we expect \(\mathbb P^n\) to classify lines in \((n+1)\)-affine space. For technical reasons (see Remark 3), we will instead view this data as classifying one-dimensional quotients of \((n+1)\)-affine space.
Example 2. Fix a field \(k\). Then a morphism \(\operatorname{Spec}k\to\mathbb P^n_{\mathbb Z}\) is basically a closed point in \(\mathbb P^n(k)\), which amounts to the data of \(n+1\) scalars \(\{a_0,\ldots,a_{n+1}\}\) which do not all vanish, up to multiplication by \(k^\times\). This is equivalent to the data of a row matrix \(k^{n+1}\to k\) of full rank, up to scalar, which is equivalent to the data of a quotient \(k^{n+1}\to k\) up to isomorphism of \(k\).
Thus, as a first approximation, we may expect that a morphism \(\operatorname{Spec}A\to\mathbb P^n_{\mathbb Z}\) has equivalent data to a quotient map \(A^{n+1}\to A\), up to isomorphism.
Remark 3. Let’s explain (in rough terms) why it is more convenient to consider quotients instead of sub-objects. The reason is that a “line” should be viewed as a subspace which admits a complement. For example, over a local ring \(A\), we should be classifying split short exact sequences \[0\to A\to A^{n+1}\to A^n\to0.\] This is a lot of data! However, note that the kernel of the map \(A^{n+1}\to A^n\) is projective and hence free (because \(A\) is local), and it then has rank one, so the isomorphism class of this short exact sequence can be remembered merely by remembering the surjection \(A^{n+1}\to A^n\). Similarly, by passing to the dual, we see that this short exact sequence has equivalent data to a surjection \(A^{n+1}\to A\).
This first approximation is wrong. In fact, there is no scheme \(P\) equipped with a natural isomorphism between the set of morphisms \(\operatorname{Spec}A\to P\) and the set of isomorphism classes of quotients \(A^{n+1}\to A\). To prove this, let’s recall some notions related to representability of functors.
Notation 4. Given a category \(\mathcal C\), we let \(\mathrm{PSh}(\mathcal C)\) denote the presheaf category of functors \(\mathcal C^{\mathrm{op}}\to\mathrm{Set}\), where morphisms are given by natural transformations. Given an object \(X\in\mathcal C\), we let \(h_X\) denote the presheaf \[h_X(Y):=\operatorname{Mor}_{\mathcal C}(Y,X).\]
Theorem 5 (Yoneda). Fix a category \(\mathcal C\). For any object \(X\) and presheaf \(\mathcal F\), there is a natural isomorphism \[\operatorname{Mor}_{\mathrm{PSh}(\mathcal C)}(h_X,\mathcal F)\cong\mathcal F(X).\]
Sketch. Given a natural transformation \(\eta\colon h_X\to\mathcal F\), we can produce the element \(\eta_X(\mathrm{id}_X)\in\mathcal F(X)\). Conversely, given an element \(x\in\mathcal F(X)\), one can construct a natural transformation \(\eta\colon h_X\to\mathcal F\) as follows: for any \(Y\in\mathcal C\) and \(f\in h_X(Y)\), we define \[\eta_Y(f):=\mathcal Ff(x).\] From here, one has to check that \(\eta\) assembles into a natural transformation and that these two constructions are inverse. One should then check that the constructions are natural in \(X\) and \(\mathcal F\). We will write out none of these checks. \(\blacksquare\)
Corollary 6. Fix a category \(\mathcal C\). Then the map \(X\mapsto h_X\) defines a fully faithful embedding \(\mathcal C\to\mathrm{PSh}(\mathcal C)\).
Proof. By Theorem 5, one has \[\operatorname{Mor}_{\mathrm{PSh}(\mathcal C)}(h_X,h_Y)\cong h_Y(X)=\operatorname{Mor}_{\mathcal C}(X,Y).\] We will be done if we can check that this map sends a morphism \(f\colon X\to Y\) to the natural transformation \((f\circ -)\colon h_X\to h_Y\). We may as well run the inverse check. To this end, by construction, \((f\circ-)\) goes to the element \(f\circ{\mathrm{id}_X}\) of \(h_Y(X)\), which is the morphism \(f\colon X\to Y\). \(\blacksquare\)
It is in general an interesting question which presheaves \(\mathcal F\) are in the image of the embedding \(\mathcal C\to\mathrm{PSh}(\mathcal C)\), in which case we may say that \(\mathcal F\) is represented by the corresponding object of \(\mathcal C\). Of course, this question is intractable in such generality, so we now return to the setting of schemes. Here is one test for representability.
Definition 7. A (Zariski) sheaf is a presheaf \(\mathcal F\in\mathrm{PSh}(\mathrm{Sch})\) satisfying the following sheaf condition: for any open cover \(\mathcal U\) of a scheme \(Y\), we have \[\mathcal F(Y)=\mathrm{eq}\left(\prod_{U\in\mathcal U}\mathcal F(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal F(U\times_YV)\right),\] where the two maps are given by the two restrictions. (Recall that \(U\times_YV\) can be identified with the intersection of the images of \(U\) and \(V\) in \(Y\).) If merely the natural map from \(\mathcal F(Y)\) is always injecive, then \(\mathcal F\) is called separated.
Remark 8. More explicitly, there is a natural map \(\mathcal F(Y)\to\prod_U\mathcal F(U)\) given by taking the product of the restrictions. By functoriality, this map factors through the equalizer to produce a map \[\mathcal F(Y)\to\mathrm{eq}\left(\prod_{U\in\mathcal U}\mathcal F(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal F(U\times_YV)\right),\] which we want to be an isomorphism.
Proposition 9. Fix a scheme \(X\). Then the presheaf \(h_X\in\mathrm{PSh}(\mathrm{Sch})\) is a Zariski sheaf.
Proof. Fix an open cover \(\mathcal U\) of a scheme \(Y\), and we are interested in showing that the natural map \[h_X(Y)\to\mathrm{eq}\left(\prod_{U\in\mathcal U}h_X(U)\rightrightarrows\prod_{U,V\in\mathcal U}h_X(U\cap V)\right)\] is an isomorphism.
We are now ready to explain why we should not ask for \(\mathbb P^n_{\mathbb Z}\) to represent the functor sending the affine scheme \(\operatorname{Spec}A\) to isomorphism classes of quotients \(A^{n+1}\to A\). By Proposition 9, it is enough to show that this is not a sheaf.
Definition 10. Let \(\mathcal Q\) denote the presheaf on \(\mathrm{Sch}\) defined by sending the scheme \(X\) to the set of isomorphism classes of quotients \(\mathcal O_X^{n+1}\to\mathcal O_X\). Here, two quotients are isomorphic if they differ by an element of \(\mathcal O_X^\times(X)\).
Remark 11. The presheaf \(\mathcal Q\) is at least separateted. Indeed, for an open cover \(\mathcal U\) of \(X\), we must show that the natural map \[\mathcal Q(X)\to\mathrm{eq}\left(\prod_{U\in\mathcal U}\mathcal Q(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal Q(U\cap V)\right)\] is injective. Choose two quotients \(\pi,\tau\colon\mathcal O_X^{n+1}\to\mathcal O_X\) which agree in the equalizer. Then each \(U\in\mathcal U\) admits some scalar \(c_U\in\mathcal O_X^\times\) for which \(\pi|_U=c_U\tau|_U\). For example, it follows that the locus where \(\pi-c\tau\) vanishes is open for each \(c\in\mathcal O_X(X)^\times\), so the locus where \(\pi-\tau\) vanishes is both open and closed. By rescaling \(\tau\), we may assume that \(\pi_x=\tau_x\) for some \(x\) in each connected component of \(X\), so it follows that \(\pi=\tau\).
Proposition 12. Suppose \(n\ge1\). Then \(\mathcal Q\) is not a sheaf, even if restricted to open covers of affine schemes by affine schemes.
Proof. In light of Remark 11, the issue will be surjectivity of the natural map. The rough idea is that one cannot in general “glue” free modules to free modules. This is an issue because we need to glue surjections \(A^{n+1}\to A\) together from an open cover, but the free module \(A\) may fail to glue! Accordingly, our counterexample will come from a projective module which fails to be free.
Let \(A\) be a Dedekind domain which is not a unique factorization domain. For example, \(A=\mathbb Z[\sqrt{-5}]\) will suffice. Then \(A\) is not a principal ideal domain, so \(A\) admits an ideal \(I\) which is not principal. Because \(I\subseteq A\), we do know that \(I\) is projective of rank at most one, and it is rank one because \(I\) is nonzero. Let \(\mathcal U\) be a trivializing open cover for \(I\); we may as well assume that each \(U\in\mathcal U\) is a distinguished affine open subset \(D(f_U)\).
Now, because \(A\) is a Dedekind domain, \(I\) can be generated by two elements, meaning that there is a surjection \(A^2\to I\). By appending by zeroes, we receive a surjection \(\pi\colon A^{n+1}\to I\). (Here is the only place where we use \(n\ge1\).) Localization is exact, so \(\pi\) induces a surjection \[A^{n+1}_f\to I_f\] for each \(f\in A\). In particular, for each \(U\in\mathcal U\), we have a surjection \(A^{n+1}_{f_U}\to I_{f_U}\cong A_{f_U}\), which we call \(\pi_U\).
To complete the proof, we will show that the tuple \((\pi_U)\) is in the codomain but not the image of the map \[\mathcal Q(A)\to\mathrm{eq}\left(\prod_{U\in\mathcal U}\mathcal Q(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal Q(U\cap V)\right).\] Here are our checks.
This section is rather technical, so I have written the rest of the post so that it is not logically necessary. On the other hand, I think it provides some explicit motivation for the presence of line bundles in Theorem 1.
In sheaf theory, the natural way to fix the problem that a presheaf is not a sheaf is to pass to the sheafification. Thus, we should sheafify \(\mathcal Q\). Of course, we should probably start by explaining what the sheafification is in our context.
Definition 13. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. A morphism \(i\colon\mathcal F\to\mathcal F^{\mathrm{sh}}\) is a sheafification if and only if \(\mathcal F^{\mathrm{sh}}\) is a Zariski sheaf, and any map from \(\mathcal F\) to a Zariski sheaf\(\mathcal G\) factors uniquely through \(i\).
There is a general process of sheafification (on a “site”), and it will be useful to us to know something about it. For motivation, if we have a separated presheaf \(\mathcal F\), the obstruction of \(\mathcal F\) being a sheaf is that there are sections which “should be there” (coming from open covers) but are not. To fix this problem, we formally add in the required sections.
Definition 14. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. For any open cover \(\mathcal U\) of a scheme \(X\), we define \[\check{\mathrm H}^0(\mathcal U;\mathcal F):=\operatorname{eq}\left(\prod_{U\in\mathcal U}\mathcal F(U)\rightrightarrows\prod_{U,U’\in\mathcal U}\mathcal F(U\times_XU’)\right).\]
This notation comes from Čech cohomology. Of course, Definition 14 is only adding in the sections from the single open cover \(\mathcal U\), but we will need to add in all sections from all open covers. To this end, we want to take a colimit over open covers.
Definition 15. Fix a scheme \(X\). Then a refinement of an open cover \(\mathcal U\) is an open cover \(\mathcal V\) of \(X\) and a function \(\alpha\colon\mathcal V\to\mathcal U\) such that \(V\subseteq\alpha(V)\) for each \(V\in\mathcal V\).
Remark 16. A refinement \(\alpha\colon\mathcal V\to\mathcal U\) induces a morphism \(\alpha^*\colon\check{\mathrm H}^0(\mathcal U;\mathcal F)\to\check{\mathrm H}^0(\mathcal V;\mathcal F)\): for example, \(\alpha\) chooses maps \(\mathcal F(\alpha(V))\to\mathcal F(U’)\) for each \(V\in\mathcal V\), which combines to a morphism \[\prod_{U\in\mathcal U}\mathcal F(U)\to\prod_{V\in\mathcal V}\mathcal F(V),\] given by the maps \(\prod_{U\in\mathcal U}\mathcal F(U)\to\mathcal F(\alpha(V))\to\mathcal F(V)\). This commutes with restriction, so we get a morphism on \(\check{\mathrm H}^0\).
Remark 17. While we’re here, we note that any two choices of refinements \(\alpha,\beta\colon\mathcal V\to\mathcal U\) actually induce the same map \(\check{\mathrm H}^0(\mathcal U;\mathcal F)\to\check{\mathrm H}^0(\mathcal V;\mathcal F)\). Indeed, for \((s_U)_U\in\check{\mathrm H}^0(\mathcal U;\mathcal F)\), we want to show that \(\alpha^*(s)=\beta^*(s)\). Well, for \(V\in\mathcal V\), we must show \((\alpha^*s)_V=(\beta^*s)_V\), which is equivalent to showing \[s_{\alpha(V)}|_V\stackrel?=s_{\beta(V)}|_V.\] But this holds because \(s_{\alpha(V)}|_{\alpha(V)\cap\beta(V)}=s_{\alpha(V)}|_{\alpha(V)\cap\beta(V)}\) (by definition of \(\mathrm{\check H}^0\)!), and \(V\subseteq\alpha(V)\cap\beta(V)\).
Definition 18. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. For a scheme \(X\), we define \[\check{\mathrm H}^0(X;\mathcal F):=\operatorname{colim}_{\mathcal U}\check{\mathrm H}^0(\mathcal U;\mathcal F),\] where the colimit is taken over open covers, with morphisms given by refinement (and Remark 16).
Remark 19. Because any two open covers admit a common refinement (and the morphisms given by refinement are canonical by Remark 17), the colimit in Definition 18 is filtered.
We now upgrade the construction \(\mathrm{\check H}^0(X;\mathcal F)\) to a presheaf.
Remark 20. Given a morphism \(f\colon Y\to X\), any open cover \(\mathcal U\) of \(X\) induces an open cover \(f^{-1}\mathcal U\) of \(Y\) by taking the pre-image. The same argument as in Remark 16 provides us with a map \(f^*\colon\check{\mathrm H}^0\left(\mathcal U;\mathcal F\right)\to\check{\mathrm H}^0\left(f^{-1}\mathcal U;\mathcal F\right)\) by using the various restrictions of \(\mathcal Ff\). Because everything is basically induced by \(\mathcal Ff\), naturality of \(\mathcal F\) makes these morphisms compatible with refinement, so we may pass to the colimit to receive a map \[f^*\colon\mathrm{\check H}^0(X;\mathcal F)\to\mathrm{\check H}^0(Y;\mathcal F).\]
Quickly, we note that the maps of Remark 20 are functorial. For example, \({\mathrm{id}_X}\) goes to the identity map on \(\mathrm{\check H}^0(X;\mathcal F)\). Also, \((f\circ g)^*=g^*\circ f^*\) for morphisms \(f\colon Y\to X\) and \(g\colon Z\to Y\). We will not write out these checks by hand, but they follow by expanding out the definition and tying everything back to the functoriality of \(\mathcal F\). We thus may make the following definition.
Definition 21. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. Then we define the presheaf \(\mathcal F^+\) by \(\mathcal F^+(X):=\mathrm{\check H}^0(X;\mathcal F)\), with morphisms as in Remark 20.
Remark 22. The construction of Remark 8 provides a map \(\mathcal F(X)\to\mathcal F^+(X)\) for any scheme \(X\). The restrictions maps of \(\mathcal F^+\) are induced by the restriction maps for \(\mathcal F\) (first on open covers \(\mathcal U\) and then by passing to the colimit), so this upgrades to a morphism \(\mathcal F\to\mathcal F^+\) of presheaves.
Remark 23. Given a morphism \(\mathcal F\to\mathcal G\) of presheaves, we get a natural map \(\mathrm{\check H}^0(\mathcal U;\mathcal F)\to\mathrm{\check H}^0(\mathcal U;\mathcal G)\) for any open cover \(\mathcal U\) by functoriality of limits. Functoriality of colimits then produces a map \(\mathcal F^+(X)\to\mathcal G^+(X)\) for any scheme \(X\). As in Remark 22, this upgrades to a morphism of presheaves by functoriality of limits and colimits.
At least in the case where \(\mathcal F\) is a separated presheaf, we expect \(\mathcal F^+\) to be the sheafification. This will turn out to be true, but it is a little tricky to check.
Lemma 24. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. For any scheme \(X\), any \(s\in\mathcal F^+(X)\) admits an open cover \(\mathcal U\) of \(X\) satisfying the following: for any \(U\in\mathcal U\), the restriction \(s|_U\) is in the image of the map \(\mathcal F(U)\to\mathcal F^+(U)\).
Proof. By definition of \(\mathcal F^+(X)\), there is an open cover \(\mathcal U\) of \(X\) so that \(s\) comes from an element \((s_U)_U\in\check{\mathrm H}^0(\mathcal U;\mathcal F)\). Now, for \(U\in\mathcal U\), we see \(s_U\in\mathcal F(U)\), which we claim agrees with the element \(s|_U\) in \(\mathcal F^+(U)\).
Well, \(s|_U\) is given by restricting the coordinates of \(s\) component-wise to the open cover \(U\cap\mathcal U\) of \(U\). The key input is that the open cover \(U\cap\mathcal U\) is refined by the open cover \(\{U\}\). Thus, \(s|_U\) agrees with the element \(s_U\in\check{\mathrm H}^0(\{U\};\mathcal F)\) once embedded into the directed limit \(\mathcal F^+(U)\). \(\blacksquare\)
Proposition 25. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. Then \(\mathcal F^+\) is separated.
Proof. Fix a scheme \(X\) and two sections \(s,s’\in\mathcal F^+(X)\) which agree when restricted to an open cover \(\mathcal U\) of \(X\). By refining \(\mathcal U\) and Lemma 24, we may assume that \(s|_U\) and \(s’|_U\) are both in the image of the natural map \(\mathcal F(U)\to\mathcal F^+(U)\) for each \(U\in\mathcal U\). We will write \(s_U\) and \(s’_U\) for these pre-images in \(\mathcal F(U)\), respectively.
Now, for each \(U\in\mathcal U\), we know that \(s_U\) and \(s’_U\) are equal in \(\mathcal F^+(U)\). This means that \(U\) admits an open cover \(\mathcal V_U\) for which \(s_U|_V=s’_U|_V\) for each \(V\in\mathcal V_U\). Then the open cover \(\mathcal V=\bigcup_{U\in\mathcal U}\mathcal V_U\) of \(X\) admits the element \[(s_U|_V)_{V\in\mathcal V_U}=(s’_U|_V)_{V\in\mathcal V_U}\] in \(\mathrm{\check H}^0(\mathcal V;\mathcal F)\) which witnesses \(s=s’\) in \(\mathcal F^+(X)\). \(\blacksquare\)
Proposition 26. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a separated presheaf. Then \(\mathcal F^+\) is a sheaf.
Proof. By Proposition 25, \(\mathcal F^+\) is already separated, so we only have to check gluing. Namely, suppose that we are given an open cover \(\mathcal U\) of a scheme \(X\) along with sections \(s_U\in\mathcal F^+(U)\) such that \(s_U|_{U\cap V}=s_V|_{U\cap V}\) for each \(U,V\in\mathcal U\). Then we would like to find \(s\in\mathcal F(X)\) restricting to each \(s_U\).
Quickly, we note that we may refine \(\mathcal U\) to some open cover \(\mathcal U’\) with no trouble. Indeed, the compatibility of the sections remains true after restriction. Then if we can find \(s\) gluing the restricted sections \(s_{U’}\) for \(U’\in\mathcal U’\), then \(s|_U|_{U\cap U’}=s_U|_{U\cap U’}\) for each \(U\in\mathcal U\) and \(U’\in\mathcal U’\), so \(s|_U=s_U\) holds for each \(U\in\mathcal U\) because \(\mathcal F^+\) is already known to be separated.
Accordingly, we use Lemma 4 to refine \(\mathcal U\) so that \(s_U\) is the image of the natural map \(\mathcal F(U)\to\mathcal F^+(U)\) for each \(U\in\mathcal U\). Because \(\mathcal F\) is separated, the natural map \(\mathcal F\to\mathcal F^+\) is injective, so this element \(s_U\) is unique. Furthermore, the equality \(s_U|_{U\cap V}=s_V|_{U\cap V}\) for each \(U,V\in\mathcal U\) takes place in \(\mathrm F^+(U\cap V)\) but upgrades to an equality in \(\mathcal F(U\cap V)\). Thus, \((s_U)\) is an element in \(\check{\mathrm H}^0(\mathcal U;\mathcal F)\), so it maps to an element \(s\) in \(\mathcal F^+(X)\). As in Lemma 24, we see that \(s|_U=s_U\), so we are done! \(\blacksquare\)
Theorem 27. Let \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) be a presheaf. Then the map \(\mathcal F\to\mathcal F^{++}\) is the sheafification of \(\mathcal F\).
Proof. By Propositions 25 and 26, \(\mathcal F^{++}\) is a sheaf. It remains to check the universal property. Well, choose a sheaf \(\mathcal G\) and a map \(\eta\colon\mathcal F\to\mathcal G\). We want to show that there is a unique induced map \(\mathcal F^{++}\to\mathcal G\).
Remark 28. The same argument shows that \(\mathcal F^+\) is the sheafification when \(\mathcal F\) is separated.
We now have a template to fix the problem presented in the previous subsection: it is impossible to represent \(\mathcal Q\) by a scheme because it is not a Zariski sheaf, but maybe the sheafification \(\mathcal Q^+\) (see Remark 28) can be represented. Line bundles arise naturally when trying to describe \(\mathcal Q^+\) explicitly, which we now explain.
Definition 29. Let \(\mathcal P\) denote the presheaf defined by sending the scheme \(X\) to the set of isomorphism classes of surjections \(\mathcal O_X^{n+1}\to\mathcal L\), where \(\mathcal L\) is a line bundle on \(X\). Two surjections \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\) and \(\tau\colon\mathcal O_X^{n+1}\to\mathcal M\) are isomorphic if and only if there is an isomorphism \(\mathcal L\cong\mathcal M\) compatible with the surjections.
Proposition 30. We have \(\mathcal Q^+\cong\mathcal P\). In particular, \(\mathcal P\) is a sheaf.
Proof. The core idea is that an element of \(\mathcal Q^+(X)\) is the data of gluing together local surjections \(\mathcal O_U^{n+1}\to\mathcal O_U\). The local \(\mathcal O_U\)s glue together to a line bundle (which is not necesssarily trivial!), and then one can glue the local surjections to a global surjection, which is the data of an element of \(\mathcal P(X)\). Making this precise is a little tricky.
Let’s start by describing a map \(\mathcal P\to\mathcal Q^+\).
Fix a scheme \(X\), and we will describe a map \(\eta\colon\mathcal P(X)\to\mathcal Q^+(X)\). Choose a surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\). Then \(\mathcal L\) admits a trivializing open cover \(\mathcal U\); choose isomorphisms \(\varphi_U\colon\mathcal L|_U\to\mathcal O_U\) for each \(U\in\mathcal U\).
We claim that the tuple \((\varphi_U\circ\pi|_U)_U\) is an element of \(\check{\mathrm H}^0(\mathcal U;\mathcal Q)\), which is then an element in \(\mathcal Q^+(X)\). Certainly each composite \(\varphi_U\circ\pi|_U\) is a surjection \(\mathcal O_U^{n+1}\to\mathcal O_U\), so we only have to show compatbility. Well, for \(U,V\in\mathcal U\), one has the composite isomorphism \[\mathcal O_U|_{U\cap V}\stackrel{\varphi_U}\cong\mathcal L|_{U\cap V}\stackrel{\varphi_V}\cong\mathcal O_V|_{U\cap V},\] which is compatible with the surjections \(\varphi_U\circ\pi|_U\) and \(\varphi_V\circ\pi|_V\) by construction, so these surjections agree when restricted to \(U\cap V\).
Continue with \(X\) fixed. We claim that the map \(\mathcal P(X)\to\mathcal Q^+(X)\) is well-defined. Changing the isomorphism class of \(\mathcal L\) does not do much to our construction, but we do have to check that changing the open cover and local trivializations does not change the target element of \(\mathcal Q(X)\).
Well, choosing a different local trivialization \(\varphi_U\colon\mathcal L|_U\to\mathcal O_U\) amounts to composing with an automorphism of \(\mathcal O_U\). This does not change the isomorphism class in \(\mathcal Q(U)\), so it does not change the underlying element of \(\mathcal Q(X)\).
Thus, we may reduce our well-definedness check to showing that the target element in \(\mathcal Q(X)\) does not change if we refine \(\mathcal U\) to some open cover \(\mathcal U’\), using the induced trivializations (by restriction). But this process of refinement restricts component-wise, so it is exactly the map \[\mathrm{\check H}^0(\mathcal U;\mathcal Q)\to\mathrm{\check H}^0(\mathcal U’;\mathcal Q),\] so it does not change the underlying class in \(\mathcal Q^+(X)\).
We now upgrade \(\eta\colon\mathcal P\to\mathcal Q^+\) to a morphism of presheaves. Well, choose a morphism \(f\colon Y\to X\) and a surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\). We may as well choose a trivializing open cover \(\mathcal U\) of \(\mathcal L\) and local trivializations \(\varphi_U\colon\mathcal L|_U\to\mathcal O_U\). Then \(\eta(f^*\pi)\) is the element \[(f^*\varphi_{f^{-1}U}\circ f^*\pi|_{f^{-1}U})_U\in\check{\mathrm H}^0\left(f^{-1}\mathcal U;\mathcal Q\right)\] because \(f^*\varphi_{f^{-1}U}\colon f^*\mathcal L\to\mathcal O_{f^{-1}U}\) are suitable local trivializations. Similarly, \(f^*\eta(\pi)\) is the element \[(f^*(\varphi_U\circ\pi|_U))_U\in\check{\mathrm H}^0\left(f^{-1}\mathcal U;\mathcal Q\right).\] These agree by funtoriality of the pullback.
We now describe the inverse map \(\mathcal Q^+\to\mathcal P\), which is trickier.
Fix a scheme \(X\), and we describe a map \(\mathcal Q^+(X)\to\mathcal P(X)\) is surjective. Choose some element in \(\mathcal Q^+(X)\), which may represented as a tuple \((\pi_U)_U\in\check{\mathrm H}^0(\mathcal U;\mathcal Q)\), where \(\mathcal U\) is some open cover of \(X\).
Let’s start by constructing our line bundle. By construction, for each \(U,V\in\mathcal U\), there is an isomorphism \(\varphi_{UV}\colon\mathcal O_V|_{U\cap V}\to\mathcal O_U|_{U\cap V}\) satisfying \(\pi_U=\varphi_{UV}\circ\pi_V\). In fact, \(\varphi_{UV}\) is unique because \(\pi_V\) is surjective and hence epic; for example, this forces \(\varphi_{UU}=\mathrm{id}\) for each \(U\). This uniqueness also automatically implies that these comparison isomorphisms satisfy the cocycle condition, so we may glue the local sheaves \(\mathcal O_U\) into a sheaf \(\mathcal L\), which is a line bundle because it is locally free. (One way to do this gluing is to first define \(\mathcal L\) as a sheaf on the base of open subsets contained in an open subset in \(\mathcal U\).)
Now, the surjections \(\pi_U\colon\mathcal O_U^{n+1}\to\mathcal O_U\) are (by construction) compatible with the isomorphisms \(\varphi_{UV}\), so they glue to a morphism \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\). (One way to see this is that the \(\pi_U\)s first define a morphism of sheaves on a base, which then uniquely extends to a morphism of sheaves.) The map \(\pi\) is locally a surjection, so it is surjective and thus provides an element of \(\mathcal P(X)\).
For fixed \(X\), and we claim that the map \(\mathcal Q^+(X)\to\mathcal P(X)\) is well-defined. Namely, we need to show that the class in \(\mathcal P(X)\) does not depend on the choice of representative \((\pi_U)\); for example, changing the isomorphism class of \(\pi_U\) does nothing to our construction except adjust the local isomorphisms throughout.
Now, by definition \(\mathcal Q^+(X)\), it is enough to check that the output in \(\mathcal P(X)\) does not change if we pass to a refinement of \((\pi_U)_{U\in\mathcal U}\), say \((\pi_{U’})_{U’\in\mathcal U’}\). For example, passing to a refinement merely restricts the comparison isomorphisms, so the same line bundle \(\mathcal L\) can be used as the outcome of the gluing. Similarly, the surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\) can be uniquely described by its restriction to any base, so the open subsets contained in an open subset of \(\mathcal U’\) will do just fine to determine it.
Lastly, we need to check that our maps are inverse. Fix a scheme \(X\).
We claim that the composite \(\mathcal P(X)\to\mathcal Q^+(X)\to\mathcal P(X)\) is the identity. Well, starting from \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\), we chose a trivializing open cover to produce an element in \(\mathcal Q^+(X)\). This trivializing open cover can then be glued straight back to the original line bundle \(\mathcal L\) and surjection, so the map \(\mathcal Q^+(X)\to\mathcal P(X)\) returns the original map \(\pi\).
We claim that the composite \(\mathcal Q^+(X)\to\mathcal P(X)\to\mathcal Q^+(X)\) is the identity. Well, starting from some tuple \((\pi_U)_{U\in\mathcal U}\), we glued these to a surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\). But \(\mathcal U\) is a trivializing open cover for \(\mathcal L\) by its construction, and the construction of \(\mathcal L\) implies that the induced local surjection \(\mathcal O_U^{n+1}\to\mathcal L|_U\cong\mathcal O_U\) (for \(U\in\mathcal U\)) is exactly \(\pi_U\). Thus, passing through the map \(\mathcal P(X)\to\mathcal Q^+(X)\) returns \((\pi_U)_U\) again. \(\blacksquare\)
Remark 31. It is not obvious from the above proof, but the sheafification map \(\mathcal Q\to\mathcal P\) is the natural embedding. Indeed, one takes a surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal O_X\) to the natural one-element tuple \((\pi)\) in \(\mathrm{\check H}^0(\{X\};\mathcal Q)\). The construction of the map \(\mathcal Q^+(X)\to\mathcal P(X)\) then allows us to take this tuple back to the surjection \(\pi\).
Remark 32. There are easier ways to show that \(\mathcal P\) is the sheafification of \(\mathcal Q\). For example, there is already a natural embedding \(\mathcal Q\subseteq\mathcal P\), and one can directly show that \(\mathcal P\) is a sheaf without too much trouble (see Lemma 33). Then one can argue as in the proof of Theorem 27 to show that \(\mathcal P\) is the sheafification, once we argue that elements of \(\mathcal P\) are locally given by elements of \(\mathcal Q\). (Namely, we should prove a version of Lemma 24 by hand.)
We promised that the previous section would not be logically necessary to our story, so let’s directly show the only piece we will need. We know that \(\mathcal Q\) is a separated presheaf, but it fails to be a sheaf. The argument of Proposition 12 suggests that \(\mathcal Q\) is missing surjections onto sheaves which are not globally free but merely locally free. This provides alternative motivation for Definition 29, which defines the presheaf \(\mathcal P\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\) by defining \(\mathcal P(X)\) to be the collection of isomorphism classes of surjections \(\mathcal O_X^{n+1}\to\mathcal L\), where \(\mathcal L\) is a line bundle on \(X\).
Lemma 33. The presheaf \(\mathcal P\) is a Zariski sheaf.
Proof. Let \(\mathcal U\) be an open cover of a scheme \(X\), so we want to show that the natural map \[\mathcal P(X)\to\operatorname{eq}\left(\prod_{U\in\mathcal U}\mathcal P(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal P(U\cap V)\right)\] is an isomorphism. The same argument as in Remark 11 shows that this map is injective, once one replaces the scalars \(c_U\) with elements of \(\operatorname{Hom}_U(\mathcal L|_U,\mathcal M|_U)\). Technically, Remark 11 only grants a global map \(\mathcal L\to\mathcal M\) by gluing, but this global map automatically upgrades to a surjection by compatibility with \(\pi\) and \(\tau\), which then automatically upgrades to an isomorphism by checking locally.
Thus, we are interested in checking the surjectivity of our natural map. We will be brief, but the moral is that line bundles can be glued into line bundles. We are given a family of surjections \(\pi_U\colon\mathcal O_U^{n+1}\to\mathcal L_U\) which are isomorphic on the intersections. The isomorphisms on intersections are compatible with the surjections, which then uniquely defines them because surjections are epimorphisms. This uniqueness can be used to show that the isomorphisms on intersections satisfy the cocycle condition, so we can glue the \(\mathcal L_U\)s into a line bundle \(\mathcal L\) on \(X\). The compatible surjections \(\pi_U\) now glue into a surjection \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\), which is the desired surjection element of \(\mathcal P(X)\). \(\blacksquare\)
Our next goal is to show that \(\mathcal P\) is represented by a scheme. As mentioned previously, it is in general difficult to prove such a result, but there are some general tools in our situation. Roughly speaking, we will be able to find an “open cover” \(\mathcal P\) by schemes, which then implies that \(\mathcal P\) is actually given by a scheme.
To make this precise, we will use the current section to say something about open covers of (pre)sheaves. To start, we should define an open embedding.
Definition 34. Fix a morphism \(\eta\colon\mathcal F\to\mathcal G\) of presheaves on a category \(\mathcal C\). Then \(\eta\) is representable if and only if the following holds: for each object \(T\in\mathcal C\) and morphism \(h_T\to\mathcal G\), the pullback \(h_T\times_{\mathcal G}\mathcal F\) is isomorphic to \(h_S\) for some \(S\in\mathcal C\). Given a class of morphisms \(\mathcal M\subseteq\operatorname{Mor}\mathcal C\) stable under isomorphisms, we say that \(\eta\) is representable by \(\mathcal M\) if and only if \(\eta\) is representable, and the induced map \[h_S\cong h_T\times_{\mathcal G}\mathcal F\to h_T\] comes from a morphism \(S\to T\) in \(\mathcal M\) (via the embedding of Corollary 6).
Example 35. Suppose the category \(\mathcal C\) admits finite limits. Given a morphism \(S\to T\) in \(\mathcal C\), we claim that \(h_S\to h_T\) is representable. Indeed, for any morphism \(h_{T’}\to h_T\), there is an induced map \(T’\to T\). The embedding of Corollary 6 preserves limits (which can be checked with some effort, using the fact that limits in \(\mathrm{PSh}(\mathcal C)\) are computed pointwise), so we conclude that \[h_{T’}\times_{h_T}h_S=h_{T’\times_TS}.\]
Example 36. Because the class of open embeddings (of schemes) is preserved by base-change, an open embedding \(U\subseteq X\) induces a morphism \(h_U\to h_X\) which is representable by open embeddings.
In light of Example 36, it makes sense to take an “open embedding of presheaves” to mean a morphism which is representable by open embeddings. Here are some additional facts about morphisms representable by open embeddings.
Lemma 37. Suppose \(\eta\colon\mathcal F\to\mathcal G\) is representable by open embeddings. Then \(\eta\) is monic.
Proof. We will show that the fiber over any \(t\in\mathcal G(T)\) in \(\mathcal F(T)\) has at most one element. Well, by the Yoneda lemma, \(t\) corresponds to a morphism \(h_T\to\mathcal G\), and an element of the fiber amounts to a lifting of this morphism to \(h_T\to\mathcal F\). Each choice of lift then uniquely induces a morphism \[h_T\to h_T\times_{\mathcal G}\mathcal F.\] (Conversely, such a morphism uniquely induces a lift \(h_T\to\mathcal F\).) But \(h_T\times_{\mathcal G}\mathcal F\) is represented by an open subscheme \(U\) of \(T\). Thus, it is now enough to note that the open embedding \(U\to T\) admits at most one section \(T\to U\). \(\blacksquare\)
Lemma 38. Suppose \(\eta\colon\mathcal F\to\mathcal G\) is representable by open embeddings. If \(\mathcal G\) is a sheaf, then \(\mathcal F\) is a sheaf.
Proof. We proceed directly. Let \(\mathcal U\) be an open cover of \(X\), and we want to show that the natural map \[\mathcal F(X)\to\operatorname{eq}\left(\prod_{U\in\mathcal U}\mathcal F(U)\rightrightarrows\prod_{U,V\in\mathcal U}\mathcal F(U\cap V)\right)\] is an isomorphism. It is at least injective because everything in sight embeds into \(\mathcal G\) by Lemma 37 (because limits commute with limits), and \(\mathcal G\) is a sheaf.
It remains to show that the natural map is surjective. Choose a tuple \((x_U)\) in the equalizer, which we would like to lift to \(\mathcal F(X)\). Because \(\mathcal G\) is a sheaf, the elements \((\eta(x_U))_{U\in\mathcal U}\) glue to an element \(y\in\mathcal G(X)\). We must show that \(y\) lifts to an element of \(\mathcal F(X)\); because \(\eta\) is monic, any lift will do.
To show the lifting, note that the Yoneda lemma lets us view \(y\) as a morphism \(h_X\to\mathcal G\). Then set \(X’\) to be the open subscheme of \(X\) representing \(h_X\times_{\mathcal G}\mathcal F\), and we would like to show that \(X’=X\). Well, for each \(U\in\mathcal U\), we know that \(y|_U\in\mathcal G(U)\) is the image of \(x_U\in\mathcal F(U)\), so the corresponding map \(y|_U\colon h_U\to\mathcal G\) factors through \(\mathcal F\) and hence through \(X’\). Thus, \(X’\) contains each \(U\in\mathcal U\), so \(X’=X\). \(\blacksquare\)
Here is our notion of an open cover.
Definition 39. Fix a presheaf \(\mathcal F\colon\mathrm{Sch}^{\mathrm{op}}\to\mathrm{Set}\). Then a collection of sub-presheaves \(\mathcal U\) is an open cover if and only if the following holds: for each scheme \(T\) and morphism \(h_T\to\mathcal F\),
- for each \(\mathcal F’\in\mathcal U\), the fiber product \(h_T\times_{\mathcal F}\mathcal F’\) is represented by an open subscheme \(T_{\mathcal F’}\subseteq T\), and
- the open subschemes \(\{T_{\mathcal F’}\}\) form an open cover of \(T\).
Example 40. An open cover \(\mathcal U\) of a scheme \(X\) induces an open cover \(\{h_U\}_U\) of \(h_X\). Indeed, pulling back by any map \(T\to X\) induces open embeddings \(U\times_XT\to T\) as in Example 36, and the disjoint union \[\bigsqcup_{U\in\mathcal U}U\times_XT\to T\] is surjective because surjections are preserved by base-change.
Morally speaking, we can glue sheaves from open covers. One cannot expect to do such gluing with presheaves (or even separated presheaves) because they are not determined locally.
Lemma 41. Fix a Zariski sheaf \(\mathcal F\) with an open cover \(\mathcal U\). Then, in the category of sheaves, \[\mathcal F=\operatorname{coeq}\left(\bigsqcup_{\mathcal F’,\mathcal F’{}’\in\mathcal U}\mathcal F’\times_{\mathcal F}\mathcal F’{}’\rightrightarrows\bigsqcup_{\mathcal F’\in\mathcal U}\mathcal F’\right).\] Here, the two maps are given by the left and right projections.
This is easy to see in the category of schemes (basically by the definition of an open cover), but it will require some work in our generality. For example, the Yoneda lemma does not preserve colimits in general.
Proof. Enumerate \(\mathcal U\) as \(\{\mathcal F_i\}_i\) for clarity. Note that there are already natural maps \(\mathcal F_i\to\mathcal F\) for each \(\mathcal F_i\to\mathcal F\), and the diagram \[\bigsqcup_{i,j}\mathcal F_i\times_{\mathcal F}\mathcal F_j\rightrightarrows\bigsqcup_{i}\mathcal F_i\to\mathcal F\] commutes by definition of the fiber product. Thus, it remains to check the universal property.
Choose a sheaf \(\mathcal G\) along with morphisms \(\varphi_i\colon\mathcal F_i\to\mathcal G\) which agree on the fiber products. We would like to show that there is a unique morphism \(\varphi\colon\mathcal F\to\mathcal G\) commuting with the \(\varphi_i\)s. To this end, choose a scheme \(X\), and we will try to determine the map \(\mathcal F(X)\to\mathcal G(X)\).
As such, let’s start with some \(x\in\mathcal F(X)\). Then we receive an induced map \(h_X\to\mathcal F\). By definition of an open cover, \(h_X\times_{\mathcal F}\mathcal F_i\) is represented by an open subscheme \(U_i\) of \(X\), and \(\{U_i\}_i\) is an open cover of \(X\). By the naturality of the Yoneda lemma, the composite \(U_i\subseteq X\to\mathcal F\) corresponds to \(x|_{U_i}\), so we see that \(x|_{U_i}\) actually lifts to an element \(x_i\in\mathcal F_i(U_i)\). The moral is that the map \(\varphi\) is required to have \[\varphi_X(x)|_{U_i}=\varphi_{U_i}(x|_{U_i})=\varphi_i(x_i)\] for each \(i\). Because \(\mathcal G\) is a sheaf (and in particular separated), this uniquely determines \(\varphi_X(x)\).
We can actally use the previous paragraph to construct \(\varphi_X\colon\mathcal F(X)\to\mathcal G(X)\). Continue with \(x\in\mathcal F(X)\). Because the elements \(x_i\) agree on projections to the fiber product (because they are just restrictions of \(x\)), the elements \(\varphi_i(x_i)\) will also agree on intersections (by hypothesis on the morphisms \(\varphi_i\)). Thus, we may glue the \(\varphi_i(x_i)\)s into a genuine element \(\varphi_X(x)\in\mathcal G(X)\).
We have thus far defined maps \(\varphi_X\colon\mathcal F(X)\to\mathcal G(X)\), but we should check naturality. Choose a morphism \(f\colon Y\to X\) and some \(x\in\mathcal F(X)\); we want to show \(\varphi(f^*x)=f^*\varphi(x)\). Let \(\{U_i\}\) be the induced open cover of \(X\) with \(x_i\in\mathcal F(U_i)\) as before. Now, for each \(i\), pulling back the composite \(U_i\subseteq X\to\mathcal F\) along \(f\), we find that \(h_Y\times_{\mathcal F}\mathcal F_i\) is represented by \(f^{-1}U_i\). Accordingly, \(f^*x|_{f^{-1}U_i}\) is represented by the element \(f^*x_i\in\mathcal F_i\left(f^{-1}U_i\right)\), and \[\varphi(f^*x)|_{f^{-1}U_i}=\varphi_i(f^*x_i)=f^*\varphi_i(x_i)=(f^*\varphi(x))|_{U_i},\] where the middle equality follows because \(\varphi_i\) is a sheaf morphism. The desired equality now follows because \(\mathcal G\) is separated. \(\blacksquare\)
Remark 42. The proof of this lemma does not use the fact that \(\mathcal F\) is a sheaf in order to construct or show uniqueness of the morphism \(\mathcal F\to\mathcal G\). (We repeatedly use that \(\mathcal G\) is separated for the uniqueness of the map, and we use the fact that \(\mathcal G\) is a sheaf once in the construction.) But of course, for \(\mathcal F\) to be the coequalizer in the category of sheaves, it must actually be a sheaf!
We are now ready to state our target tool.
Theorem 43. Let \(\mathcal F\) be a Zariski sheaf. Suppose that \(\mathcal F\) admits an open cover \(\mathcal U\), and each \(\mathcal F’\in\mathcal U\) is represented by a scheme. Then \(\mathcal F\) is represented by a scheme.
Intuitively, sheaves should be determined by their local properties, so we are saying that a sheaf which is “locally” a scheme can glue this property to make the original sheaf “globally” a scheme.
Proof. The main point is the constuction of the scheme. Enumerate \(\mathcal U\) as \(\{\mathcal F_i\}_i\) for clarity, and for each \(i\), let \(U_i\) represent \(\mathcal F_i\). By Lemma 41, we basically want to find a scheme \(X\) which admits an open cover by \(\{U_i\}\) with the “same” intersections. Thus, we need to do some gluing, which we do in two steps.
We construct the gluing data. By definition of the open cover, the map \(\mathcal F_i\times_{\mathcal F}\mathcal F_j\to\mathcal F_i\) is represented by some open subscheme \(U_{ij}\subseteq U_i\). Then for each \(i\) and \(j\), there is a canonical identification \[U_{ij}\stackrel{\mathrm{pr}_i}\cong\mathcal F_i\times_{\mathcal F}\mathcal F_j\stackrel{\mathrm{pr}_j}\cong U_{ji},\] which we call \(\varphi_{ji}\). (In other words, \(\varphi_{ji}\colon U_{ij}\to U_j\) is induced by a projection.) For example, \(U_{ii}=U_i\) and \(\varphi_{ii}=\mathrm{id}\) by construction.
To glue these open subschemes along these intersections, we must check a cocycle condition. This is basically automatic. Fix indices \(i\), \(j\), and \(k\). Then each of \(U_{ij}\cap U_{ik}\) and \(U_{ji}\cap U_{jk}\) and \(U_{ki}\cap U_{kj}\) is canonically isomorphic to \(\mathcal F_i\times_{\mathcal F}\mathcal F_j\times_{\mathcal F}\mathcal F_k\) via the various projections. For example, \[\begin{align}U_{ij}\cap U_{ik} &= U_{ij}\times_{U_i}\times U_{ik} \\ &= (\mathcal F_i\times_{\mathcal F}\mathcal F_j)\times_{\mathcal F_i}(\mathcal F_i\times_{\mathcal F}\mathcal F_k) \\ &= \mathcal F_i\times_{\mathcal F}\mathcal F_j\times_{\mathcal F}\mathcal F_k,\end{align}\] where the last isomorphism follows by formal properties of the fiber product. (Notably, the second isomorphism is given by projecting along the isomorphism \(\mathcal F_i\to U_i\), so we can see the composite isomorphism from the bottom to the top is given by this projection.) From here, the cocycle condition follows because the various comparison isomorphisms between our three open sets are all compatible with the canonical isomorphism to \(\mathcal F_i\times_{\mathcal F}\mathcal F_j\times_{\mathcal F}\mathcal F_k\) (as we just checked).
Thus, we have a scheme \(X\) which admits an open cover by \(\{U_i\}\), and the intersection inclusion \(U_i\cap U_j\subseteq U_i\) is isomorphic to the inclusion \(U_{ij}\subseteq U_i\), which is isomorphic to the projection \(\mathcal F_i\times_{\mathcal F}\mathcal F_j\to\mathcal F_i\). Thus, the coequalizer diagram \[\bigsqcup_{i,j}U_i\cap U_j\rightrightarrows\bigsqcup_iU_i\to X\] of Lemma 41 (see Example 40) is isomorphic to the coequalizer diagram \[\bigsqcup_{i,j}\mathcal F_i\times_{\mathcal F}\mathcal F_j\rightrightarrows\bigsqcup_i\mathcal F_i\to\mathcal F\] of Lemma 41 (again). It follows that \(h_X\cong\mathcal F\), and we are done! \(\blacksquare\)
Remark 44. We only used the hypothesis that \(\mathcal F\) is a sheaf when applying Lemma 41. In particular, the construction of the candidate scheme \(X\) did not use the sheaf condition.
Remark 45. The proof of Theorem 43 is “effective” in the sense that it explicitly constructs the representing scheme (by gluing).
We are now ready to show that there is a scheme representing our sheaf \(\mathcal P\).
Proposition 46. The sheaf \(\mathcal P\) is representable.
Proof. Recall that \(\mathcal P\) is a sheaf by Proposition 30 or Lemma 33. Thus, by Theorem 43, it is enough to show that \(\mathcal P\) admits an open cover by representable (open) subsheaves.
To this end, fix an index \(i\in\{0,1,\ldots,n+1\}\) for now, and define the presheaf \(\mathcal P_i\) to take a scheme \(X\) to the set of surjections \(\pi\in\mathcal P(X)\) such that the composite \[\mathcal O_X\stackrel{\iota_i}\hookrightarrow\mathcal O_X^{n+1}\stackrel\pi\to\mathcal L\] is surjective, where the left map is the inclusion into the \(i\)th coordinate. Note that the \(i\)th coordinate being surjective is preserved by pullback, so \(\mathcal P_i\) is in fact a sub-presheaf.
To apply Theorem 43, we now must show that each \(\mathcal P_i\) is representable and that they assemble into an open cover of \(\mathcal P\).
Let’s start by checking that \(\mathcal P_i\) is representable, where \(i\) is still fixed. For ease of notation, take \(i=0\), and we claim that \(\mathcal P_i\) is represented by the affine scheme \(\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n]\). Indeed, for any scheme \(X\), the universal property of affine schemes gives \[\operatorname{Mor}_{\mathrm{Sch}}(X,\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n])=\operatorname{Hom}(\mathbb Z[x_1,\ldots,x_n],\mathcal O_X(X)),\] which is then equivalent to the data of \(n\) global sections \(\{s_1,\ldots,s_n\}\) of \(\mathcal O_X\) (by taking the image of the \(x_i\)s). This then produces the surjection \((1,s_1,\ldots,s_n)\colon\mathcal O_X^{n+1}\to\mathcal O_X\) in \(\mathcal P_i(X)\). Here are our checks on this construction.
These maps \(\eta_X\colon h_{\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n]}(X)\to\mathcal P_i(X)\) are functorial in the scheme \(X\). Indeed, choose a morphism \(f\colon Y\to X\) and \((s_1,\ldots,s_n)\in h_{\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n]}(X)\). On one hand, \(\eta(f^*(s_1,\ldots,s_n))\) is the surjection \((1,f^\sharp s_1,\ldots,f^\sharp s_n)\). On the other hand, \(f^*\eta(s_1,\ldots,s_n)\) is the surjection \(f^*(1,s_1,\ldots,s_n)\). These agree by construction of the pullback.
The maps \(\eta_X\colon h_{\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n]}(X)\to\mathcal P_i(X)\) are injective. Indeed, if the two surjections \((1,s_1,\ldots,s_n)\) and \((1,t_1,\ldots,t_n)\) are isomorphic, then there is an isomorphism \(\sigma\colon\mathcal O_X\to\mathcal O_X\) for which \((1,s_1,\ldots,s_n)=\sigma\circ(1,t_1,\ldots,t_n)\). The first coordinate forces \(\sigma=1\), so the two surjections are actually equal.
The maps \(\eta_X\colon h_{\operatorname{Spec}\mathbb Z[x_1,\ldots,x_n]}(X)\to\mathcal P_i(X)\) are surjective. This is a little trickier. Choose some \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\) in \(\mathcal P_i(X)\). The main claim is that \(\pi\circ\iota_i\) is an isomorphism: we may check this on affines, where \(\pi\circ\iota_i\) is a surjection of projective modules of rank one over a local ring. The kernel of this map is also projective, but it must have rank zero, so it vanishes. Thus, \(\pi\circ\iota_i\) is injective (on affines) and hence an isomorphism.
To use the main cain claim, we adjust \(\pi\) by the isomorphism \(\pi\circ\iota_i\colon\mathcal O_X\to\mathcal L\). Thus, we may assume that \(\mathcal L=\mathcal O_X\) and that \(\pi\circ\iota_i\) is the identity. The remaining coordinates of \(\pi\colon\mathcal O_X^{n+1}\to\mathcal O_X\) are given by some choice of global section, so \(\pi\) now takes the form \((1,s_1,\ldots,s_n)\) for some global sections \(\{s_1,\ldots,s_n\}\).
Continue with \(i\) fixed. We claim that the map \(\mathcal P_i\to\mathcal P\) is representable by open embeddings. Indeed, choose a test scheme \(X\) and a map \(h_X\to\mathcal P\), which in turn corresponds to some \(\pi\colon\mathcal O_X^{n+1}\to\mathcal L\) in \(\mathcal P(X)\). Let \(U_i\subseteq X\) be the subspace of \(p\in X\) where \(\pi\circ\iota_i\) is surjective at \(p\). We will show that \(U_i\) is open and that \(U_i=X\times_{\mathcal P}\mathcal P_i\), which will complete the proof of the claim.
Let’s check that \(U_i\) is open. Choose some \(p\in U_i\), and we need to find an open neighborhood of \(p\) contained in \(U_i\). This is a local problem, so we may shrink \(X\). In paritcular, we may assume that \(X\) is affine—say \(X=\operatorname{Spec}A\)—and that \(\mathcal L\) is trivial, and we go ahead and identify \(\mathcal L\) with \(\mathcal O_X\). Then \(\pi\) corresponds to a surjective map \(A^{n+1}\to A\), and the composite \(\pi\circ\iota_i\) is surjective at the prime \(p\).
Now, let \(e_i\in A^{n+1}\) be the \(i\)th basis vector, so we are given tha \(\pi(e_i)\) is a unit in \(A_p\). In particular, \(\pi(e_i)\in A\) is not a zero-divisor, so we may use the open subset \(D(\pi(e_i))\subseteq\operatorname{Spec}A\): for each \(q\in D(\pi(e_i))\) (such as \(p\)), we know that \(\pi(e_i)\) is a unit in \(A_q\), so \((\pi\circ\iota_i)_q\) is surjective, so \(q\in U_i\).
We check that \(U_i=X\times_{\mathcal P}\mathcal P_i\). By its construction as a union of distinguished affine open subsets in the previous point, \(\pi|_{U_i}\in\mathcal P_i(U_i)\), which provides the needed projection \(U_i\to\mathcal P_i\). Indeed, choose a test scheme \(T\) equipped with maps \(t\colon T\to X\) and \(T\to\mathcal P_i\) which agree in \(\mathcal P\). We would like to show that \(t\colon T\to X\) factors through \(U_i\): the factorization is unique because the map \(U_i\subseteq X\) is monic, and the composite \(T\to U_i\to\mathcal P_i\) must be the given map \(T\to\mathcal P_i\) because the map \(\mathcal P_i\to\mathcal P\) is also monic.
Now, to show that \(t\colon T\to X\) factors through \(U_i\), we should show that \(t(p)\in U_i\) for each \(p\), meaning that \((\pi\circ\iota_i)_{t(p)}\) is surjective. Well, the map \(T\to\mathcal P\) is given by the surjection \(t^*\pi\in\mathcal P(T)\), which we know is actually in the image of \(\mathcal P_i(T)\). Thus, \(t^*\pi\circ\iota_i\) is surjective, so it is surjective at every point, so \(\operatorname{im} t\subseteq U_i\) follows.
Lastly, we need to check that the maps \(\mathcal P_i\to\mathcal P\) assemble into an open cover as \(i\) varies. We will have to use Nakayama’s lemma. Choose a scheme \(X\) along with a morphism \(\pi\colon h_X\to\mathcal P\), and then define the open subscheme \(U_i\subseteq X\) for each \(i\) as in the previous check. We must show that \(\{U_i\}_i\) covers \(X\).
In other words, for each point \(p\in X\), we must show that the surjection \(\pi_p\colon\mathcal O_p^{n+1}\to\mathcal L_p\) is surjective on some coordinate. Now, let \(\{e_0,\ldots,e_n\}\) be the basis vectors of \(\mathcal O_p^{n+1}\). Then the elements \[\{\pi(e_0),\ldots,\pi(e_n)\}\] generate \(\mathcal L\), so they span \(\mathcal L_p/\mathfrak m_p\mathcal L_p\). The latter module is a vector space (over \(\mathcal O_p/\mathfrak m_p\)) of dimension one, so one of the elements \(\pi(e_i)\) forms a basis. By Nakayama’s lemma (!), this element lifts to a generator of \(\mathcal L_p\), which means that \(\pi\circ\iota_i\) is surjective at \(p\), as desired. \(\blacksquare\)
We could now simply define \(\mathbb P^n_{\mathbb Z}\) to be the scheme constructed by Proposition 46. This then makes the proof of Theorem 1 obvious, once we remember that the data of \(n+1\) global sections of a line bundle \(\mathcal L\) is the data of a map \(\mathcal O_X^{n+1}\to\mathcal L\), and saying that the global sections have no common zero is the same as saying that the map is surjective (by Nakayama’s lemma).
Of course, this is not an honest proof because \(\mathbb P^n_{\mathbb Z}\) admits a more classical definition by gluing affines. Let’s compare these two definitions of \(\mathbb P^n_{\mathbb Z}\).
Proof of Theorem 1. Let \(X\) be the scheme constructed from Proposition 46, which we would like to show is isomorphic to \(\mathbb P^n_{\mathbb Z}\). We will use Theorem 43 to describe \(X\) explicitly by gluing.
For each \(i\in\{0,1,\ldots,n+1\}\), we have an open subscheme \(U_i\subseteq X\) isomorphic to \(\mathbb A^n_{\mathbb Z}\). We write \(U_i\) as \(\operatorname{Spec}\mathbb Z[x_{0/i},\ldots,x_{n/i}]\) for psychological reasons, where \(x_{i/i}\) is understood to be \(1\). By unwinding the proof of representability in Proposition 46, we find that the map \(U_i\to X\) corresponds to the surjection \[(x_{0/i},\ldots,x_{n/i})\colon\mathcal O_{U_i}^{n+1}\to\mathcal O_{U_i},\] where we recall \(x_{i/i}=1\).
Next, let’s compute the intersections. As in the proof of Proposition 46, the intersection \(U_{ij}=U_i\times_{X}U_j\) embedded in \(U_i\) is exactly the locus where the surjection \((x_{0/i},\ldots,x_{n/i})\) is surjective in the \(j\)th coordinate. This is equivalent to \(x_{j/i}\) being a unit, so we conclude that \[U_{ij}=D(x_{j/i})=\operatorname{Spec}\mathbb Z[x_{0/i},\ldots,x_{n/i},1/x_{j/i}].\]
Lastly, we compute the transition maps. Namely, we should compute the map \(U_{ij}\to U_j\), which we know is an isomorphism onto \(U_{ji}\). To this end, we note that the surjection \((x_{0/i},\ldots,x_{n/i})\) is isomorphic to the surjection \((x_{0/i}/x_{j/i},\ldots,x_{n/i}/x_{j/i})\) over \(U_{ij}\). By the proof of Proposition 46, we see that this surjection has its \(j\)th coordinate equal to one and thus corresponds to the ring map \[\mathbb Z[x_{0/j},\ldots,x_{n/j}]\to\mathbb Z[x_{0/i},\ldots,x_{n/i},1/x_{j/i}]\] defined by \(x_{k/j}\mapsto x_{k/i}/x_{j/i}\), which in turn defines the desired transition map \(U_{ij}\to U_j\).
The above gluing data is exactly the usual construction of \(\mathbb P^n_{\mathbb Z}\)! Thus, \(X\cong\mathbb P^n_{\mathbb Z}\). \(\blacksquare\)
One indication that we have built a robust theory is that it is now not so hard to prove related results. We close the post with sketches of two such applications.
Remark 47. One can discuss Grassmannians using essentially the same arguments. For positive integers \(d\) and \(n\) with \(d\le n\), here are the relevant statements.
- The sheafification of the presheaf taking a scheme \(X\) to the set of isomorphism classes of surjections \(\mathcal O_X^n\to\mathcal O_X^d\) is the sheaf \(\operatorname{Gr}_{n,d}\) taking a scheme \(X\) to the set of isomorphism classes of surjections \(\mathcal O_X^n\to\mathcal E\), where \(\mathcal E\) is a vector bundle on \(X\) of rank \(d\).
- The sheaf \(\operatorname{Gr}_{n,d}\) is representable. In fact, it admits an open covering by the affine schemes \(\mathbb A_{\mathbb Z}^{d(n-d)}\).
]]>Remark 48. Fix a vector bundle \(\mathcal E\) on a scheme \(S\), and consider the presheaf \(\mathbb P\mathcal E\) taking a scheme \(X\) to the data of a morphism \(f\colon X\to S\), a line bundle \(\mathcal L\) on \(X\), and a surjection \(\pi\colon f^*\mathcal E\to\mathcal L\) (up to isomorphism). Then \(\mathbb P\mathcal E\) is representable.
Here is a sketch of an argument. One can adapt Lemma 33 to show that \(\mathbb P\mathcal E\) is a sheaf, so by Theorem 43, it remains to find a suitable open cover. Choose a trivializing open cover \(\mathcal U\) of \(S\) for \(\mathcal E\). Then for each \(U\in\mathcal U\), there is a natural map \[\mathbb P^{\operatorname{rank}\mathcal E}_{\mathbb Z}\times U\cong\mathbb P(\mathcal E|_U)\to\mathbb P\mathcal E\] representable by open embeddings, and these maps form an open cover of \(\mathbb P\mathcal E\).
Our arguments will follow those in Vakil. Throughout, a variety is a reduced separated scheme of finite type over a field, and a curve is a one-dimensional variety. Here is today’s target.
Theorem 1. Fix a field \(k\). Then the following categories are equivalent.
- The category of regular projective integral curves over \(k\), with morphisms given by surjections.
- The category of integral curves over \(k\), with morphisms given by dominant rational maps.
- The opposite category of finitely generated field extensions of \(k\) of transcendence degree \(1\).
The equivalence of the bottom two categories is rather formal.
Proposition 2. Fix a field \(k\). Then the following categories are equivalent.
- The category of integral varieties over \(k\), with morphisms given by dominant rational maps.
- The opposite category of finitely generated field extensions of \(k\).
Furthermore, the equivalence sends varieties of dimension \(d\) to field extensions of transcendence degree \(d\).
Proof. Given an integral variety \(X\), then the stalk \(K(X)\) at the generic point is a field extension. Furthermore, a dominant rational map \(f\colon X\to Y\) necessarily maps the generic point to the generic point, so we induce a map \(f^\sharp\colon K(Y)\to K(X)\). Taking stalks is functorial, so we have defined a functor from the first point to the second. Here are our checks on this functor.
The above checks show that we have an equivalence. The last sentence of the proposition follows because the dimension of an integral variety equals the dimension of its fraction field, and the dimension of the fraction field equals the transcendence degree. \(\blacksquare\)
The equivalence of the first two categories in Theorem 1 will require more work. There is certainly an inclusion from the first category to the second, so the difficulty lies in checking that it has good properties. In particular, it is not obvious that the inclusion is full or essentially surjective.
For example, for essential surjectivity, we need to be able to produce regular curves, so it will be useful to have an easy test for regularity. However, it is not too hard to achieve normality by simply passing to the normalization. Accordingly, we will use the following piece of commutative algebra.
Lemma 3. Fix a Noetherian local domain \((A,\mathfrak m)\) of dimension one. Then the following are equivalent.
- \(A\) is a discrete valuation ring.
- \(A\) is normal.
- \(\mathfrak m\) is principal.
- \(A\) is regular.
Proof. We show the implications in sequence. This proof is long but only because there are many things to show.
We show 1 implies 2. Suppose \(A\) is a discrete valuation ring with valuation \(v\). Rescale the valuation so that \(v(A)=\mathbb Z\). Then we claim that \(A\) is a unique factorization domain, which will complete the proof. Choose a uniformizer \(\varpi\) with \(v(\varpi)=1\). Then any \(a\in A\) either has \(a/\varpi\in A\) or \(v(a)=0\) (and so \(a\in A^\times\)), so \((\varpi)\) is the unique maximal ideal of \(A\), so \(\varpi\) is prime. Now, any \(a\in A\) can be written as \[a=u\varpi^n\] where \(n=v(a)\), which provides our factorization. For uniqueness, note that \(u\varpi^n=u’\varpi^{n’}\) implies \(n=n’\) by taking valuations and so \(u=u’\) by dividing.
We show 2 implies 3. This is pretty tricky. Choose any nonzero \(a\in\mathfrak m\), which we would like to normalize into a generator of \(\mathfrak m\). Then \(A/(a)\) is a zero-dimensional Noetherian local ring with maximal ideal \(\mathfrak m/(a)\). Thus, \(\mathfrak m/(a)\) is the only prime, so it is the nilradical, so some power vanishes. Let \(N\) be the least nonnegative integer for which \(\mathfrak m^N\subseteq(a)\). Note \(N>0\) because \(a\) is not a unit.
Now, choose any \(b\in\mathfrak m^{N-1}\setminus(a)\), and we will show that \(a/b\) generates \(\mathfrak m\). It is enough to show \((b/a)\mathfrak m=A\). On one hand, \(b\mathfrak m\subseteq\mathfrak m^N\subseteq(a)\), so \((b/a)\mathfrak m\subseteq A\).
For the other inclusion, note \((b/a)\mathfrak m\) is an ideal of \(A\), so it is enough to show that \((b/a)\mathfrak m\not\subseteq\mathfrak m\). Well, \(b\notin(a)\) means \(b/a\notin A\), so \(b/a\) cannot even be integral over \(A\) by normality! But \((b/a)\mathfrak m\subseteq\mathfrak m\) would imply that \(\mathfrak m\) is a faithful, finitely generated module over \(A[b/a]\), forcing \(b/a\) to be integral, which is a contradiction.
We show 3 and 4 are equivalent. Note \(A\) is regular so that \(\dim_{A/\mathfrak m}\mathfrak m/\mathfrak m^2=1\). So if \(\mathfrak m\) is principal, generated by \((\varpi)\), then \(\mathfrak m/\mathfrak m^2\) is spanned by \((\varpi)\) and is so one-dimensional. The converse follows from Nakayama’s lemma, which explains that a basis of \(\mathfrak m/\mathfrak m^2\) lifts to a set of generators.
We show 3 implies 1. We would like to use the valuation \[v(a):=\max\{n\in\mathbb N:a\in\mathfrak m^n\}.\] Quickly, note that \(\bigcap_{n}\mathfrak m^n=0\) by Nakayama’s lemma (because this ideal multiplied by \(\mathfrak m\) is itself); thus, \(v(a)\) is finite when \(a\) is nonzero.
To check that \(v\) has good properties, we claim that \((a)=\mathfrak m^{v(a)}\). This immediately implies that \(v(a+b)\ge\min\{v(a),v(b)\}\) and \(v(ab)=v(a)+v(b)\) for \(a,b\in A\), thereby completing the proof.
It remains to show the claim. Certainly \(a\in\mathfrak m^{v(a)}\) by construction, so \((a)\subseteq\mathfrak m^{v(a)}\). For the converse, choose a generator \(\varpi\) of \(\mathfrak m\). Then \(a\in\mathfrak m^{v(a)}\) implies that \(a/\varpi^{v(a)}\in A\), but \(a\notin\mathfrak m^{v(a)+1}\) implies that \(a/\varpi^{v(a)}\notin\mathfrak m\). Thus, \(a/\varpi^{v(a)}\in A^\times\), so \((a)=(\varpi)^{v(a)}\), and the claim follows. \(\blacksquare\)
This is all that we will say about the essential surjectivity check for now. For the fullness check, we basically need to check that one can extend rational maps to genuine morphisms. Over curves, this is the following piece of algebraic geometry.
Proposition 4. Fix a base field \(k\). Let \(C\) be a curve, and let \(Y\) be a projective variety. For any normal point \(x\in C\), any morphism \(C\setminus\{x\}\to Y\) extends to a morphism \(C\to Y\).
Proof. This follows from the valuative criteria of properness, but we will give a direct argument. The idea is to arrange ourselves into a position where we can “clear denominators” in the definition of our morphism \(\varphi\colon C\setminus\{x\}\to Y\). Accordingly, we have many reductions.
We now complete the proof. By the last step, the sections \(\{s_0,\ldots,s_n\}\) are some regular functions on \(C\), and they fail to vanish simultaneously on \(C\setminus\{p\}\). We can now divide out these sections by \[t^{\min_i\{v_x(s_i)\}}.\] This does not change the subspace spanned by these sections over \(C\setminus\{p\}\) because \(t\) is a unit there, so we have not changed the morphism \(\varphi\). However, there is not a section \(s_i\) with \(v_x(s_i)=0\), so the sections do not simultaneously vanish at \(x\), so \(\varphi\) extends to a map on \(C\)! \(\blacksquare\)
We are now ready to prove Theorem 1.
Proof of Theorem 1. The bottom two categories are equivalent by Proposition 2, so it is enough to show that the top two categories are equivalent. There is a natural inclusion functor, which we would like to show is an equivalence.
Faithful: if two surjective morphisms \(f,g\colon X\to Y\) agree as rational maps, then they agree on an open subset. But these are varieties, so they locus where they agree is a closed subset of \(X\), so \(f=g\) follows.
Full: fix a dominant rational map \(f\colon X\to Y\) of regular projective curves. We would like to show that \(f\) extends to a surjective morphism. Well, \(f\) is already defined over an open subset \(U\subseteq X\), which is cofinite because \(X\) is a curve. Thus, one can inductively use Proposition 4 to extend \(f\) to a dominant morphism \(X\to Y\). (Notably, all points are regular and hence normal by Lemma 3.) Lastly, \(f\) is surjective because \(X\) is projective and hence proper, which implies that the image of \(f\) must be closed.
Essentially surjective: we need to show that any integral curve \(C\) is birational to a smooth projective one. We start by passing to an affine open subset of \(C\), which means that there is a closed embedding \(C\subseteq\mathbb A^n\). Thus, there is a locally closed embedding \(C\subseteq\mathbb P^n\). Then \(C\) is open in its closure, so we may pass from \(C\) to the closure, making \(C\) projective.
Lastly, let \(X\) be the normalization of \(C\). On affine open subschemes, this process is given by taking the normalization in the fraction field, so \(X\to C\) does not change the fraction field, so \(X\to C\) is birational by Proposition 2. Furthermore, \(X\to C\) is finite and hence projective, so \(X\) is projective. Lastly, \(X\) being normal implies that \(X\) is regular by Lemma 3. \(\blacksquare\)
Theorem 1 (Maschke). Fix a finite group \(G\). Then any finite-dimensional complex representation of \(G\) is isomorphic to a direct sum of irreducible representations of \(G\).
Remark 2. In fact, the hypothesis that the representations are finite-dimensional is unnecessary, and we will remove it in Corollary 11 with the help of transfinite induction.
Remark 3. Our proof will work for representations of \(G\) over any field \(k\) of characteristic not dividing \(\left|G\right|\). Slight modifications will also work for continuous representations of compact groups.
I am aware of approximately two (related) approaches to Theorem 1.
We will more or less follow the first proof, but we will attempt to do as little calculation as possible. The rough sketch is that the Reynolds operator can be used to show that the functor \((-)^G\) is exact. (See Proposition 9 and Corollary 10.) However, it will take some time to explain why this is relevant.
Let’s begin by recalling some properties of projective modules. In our application, we will take \(R\) to be the ring \(\mathbb C[G]\) so that a representation of \(G\) is just \(\mathbb C[G]\)-module.
Lemma 4. Let \(R\) be a ring, and choose some \(R\)-module \(P\). The following are equivalent.
- The functor \(\operatorname{Hom}_R(P,-)\) is exact.
- Any short exact sequence \[0\to A\to B\to P\to0\] splits.
- The module \(P\) is a direct summand of a free module.
A module satisfying any of these conditions is called projective.
Proof. We show the two implications separately.
To show that 3 implies 1, we recall that \(\operatorname{Hom}_R(P,-)\) is a right adjoint to \(-\otimes_R P\), so it is always left exact. It only remains to show that a surjection \(p\colon M\to M’’\) induces a surjection \(\operatorname{Hom}_R(P,M)\to\operatorname{Hom}_R(P,M’’)\). To this end, choose some map \(b\colon P\to M’’\), which we would like to lift to a map \(P\to M\).
We now use 3 to write \(P\oplus Q=F\) for some other module \(Q\) and free module \(F\). Because \(F\) is free, one can lift the composite \(F\to P\to M’’\) to a map \(F\to M\): indeed, simply choose lifts in \(M\) for the image of each generator of \(F\). Then the desired lift \(P\to M\) is the composite \(P\to F\to M\). \(\blacksquare\)
Roughly speaking, to show Theorem 1, we see that it will be enough to show that any short exact sequence splits, so we should show that every module is projective. Accordingly, we should show that the functor \(\operatorname{Hom}_{\mathbb C[G]}(V,-)\) is always exact. We will do this in two steps.
Lemma 5. Let \(V\) and \(W\) be complex representations of a finite group \(G\). Then \[\operatorname{Hom}_G(V,W)=\operatorname{Hom}_{\mathbb C}(V,W)^G.\] Here, \((-)^G\colon\operatorname{Rep}_{\mathbb C}(G)\to\mathrm{Vec}_{\mathbb C}\) is the invariants functor given by \(V^G=\{v\in V:gv=v\text{ for all }g\in G\}\).
Proof. This follows more or less from the definition of the \(G\)-action on \(\operatorname{Hom}_{\mathbb C}(V,W)\), which we recall is given by \[(g\varphi)(v)=g\cdot\varphi\left(g^{-1}\cdot v\right),\] for \(g\in G\) and \(\varphi\in\operatorname{Hom}_{\mathbb C}(V,W)\) and \(v\in V\). Thus, we see that a linear map \(\varphi\colon V\to W\) preserves the \(G\)-action if and only if \(g\cdot\varphi(v)=\varphi(g\cdot v)\) for all \(v\in V\) and \(g\in G\), which is equivalent to \(g\cdot\varphi=\varphi\) for all \(g\in G\). \(\blacksquare\)
Thus, to show that \(\operatorname{Hom}_G(V,-)\) is exact, we may as well show that the functors \(\operatorname{Hom}_{\mathbb C}(V,-)\) and \((-)^G\) are exact. The former is not so hard because all vector spaces are free modules. For the latter, we will require a trick.
Remark 6. To slightly motivate what follows (in Remark 8), note that Lemma 5 implies that the functor \((-)^G\) is naturally isomorphic to the functor \(\operatorname{Hom}_G(\mathbb C,-)\), where \(\mathbb C\) refers to the trivial representation.
Definition 7. Fix a finite group \(G\). Then we define the coinvariants functor \((-)_G\colon\operatorname{Rep}_{\mathbb C}(G)\to\mathrm{Vec}_{\mathbb C}\) by \[V_G:=V/I_GV,\] where \(I_G\subseteq\mathbb C[G]\) is the ideal generated by the elements \(\{g-1\}_{g\in G}\). Equivalently, \(I_GV\subseteq V\) is the subspace \(\{gv-v\}_{g\in G,v\in V}\).
Remark 8. We claim that \((-)_G\) is naturally isomorphic to the functor \(\mathbb C\otimes_{\mathbb C[G]}-\), where \(\mathbb C\) refers to the trivial representation. Let’s construct the two inverse natural isomorphisms.
- For any representation \(V\), there is a natural quotient map \(V\to\mathbb C\otimes_{\mathbb C[G]}V\) given by \(v\mapsto1\otimes v\). This map contains \((g-1)v\) in its kernel for any \(g\in G\) and \(v\in V\), so \(I_GV\) is in the kernel.
- For any representation \(V\), there is a natural \(\mathbb C\)-bilinear map \(\mathbb C\times V\to V\) given by scalar multiplication. Taking a further quotient to \(V_G\) produces a \(\mathbb C[G]\)-bilinear map because \(g\overline v=\overline v\) for any \(g\in G\) and \(\overline v\in V_G\), so we receive a natural map \(\mathbb C\otimes_{\mathbb C[G]}V\to V_G\).
Remark 8 in particular implies that the functor \((-)_G\) is right exact because it is a left adjoint. The relevance of this functor is the following.
Proposition 9. Fix a finite group \(G\). The functors \((-)^G\) and \((-)_G\) are naturally isomorphic.
Proof. The natural transformation \((-)^G\Rightarrow(-)_G\) can be defined directly: for a representation \(V\), and then we define the natural map \(V^G\to V_G\) as the composite \[V^G\subseteq V\twoheadrightarrow V_G.\] The difficulty lies in constructing an inverse, for which we use the Reynolds operator: for any representation \(V\), note that there is a map \(R_V\colon V\to V\) given by \[R_V(v)=\frac1{\left|G\right|}\sum_{g\in G}gv.\] Here are some checks.
Thus, we have constructed a natural transformation \(R\colon(-)_G\to(-)^G\). Here are the inverse checks; fix a representation \(V\).
Corollary 10. Fix a finite group \(G\). Then the functor \((-)^G\) is exact.
Proof. Note that \((-)^G\) is a right adjoint by Remark 6, so it is left exact. Similarly, \((-)_G\) is a left adjoint by Remark 8, so it is right exact. We conclude by Proposition 9. \(\blacksquare\)
Let’s collect everything we’ve said.
Proof of Theorem 1. Thus far, we have managed to show that all representations are projective modules: indeed, for any representation \(V\), we should show that the functor \(\operatorname{Hom}_G(V,-)\) is exact. By Lemma 5, it is enough to show that the two functors \(\operatorname{Hom}_{\mathbb C}(V,-)\) and \((-)^G\) are exact. The former holds by Lemma 4 because \(V\) is free over \(\mathbb C\), and the latter holds by Corollary 10.
We now turn to the proof of Theorem 1. Say that a representation \(V\) is reducible if and only if it is isomorphic to a direct sum of irreducible representations. We would like to show that all representations are irreducible.
We will show that all finite-dimensional representations are reducible by induction on dimension. The zero-dimensional case has no content because such a representation is the empty sum of irreducible representations. For the induction, note that a representation \(V\) is either irreducible, in which case we are done, or \(V\) admits a proper nonzero subrepresentation \(W\). For the latter case, note that the short exact sequence \[0\to W\to V\to V/W\to0\] splits by Lemma 4, so the reducibility of \(V\) follows because the inductive hypothesis grants the reducibility of \(W\) and \(V/W\). \(\blacksquare\)
While we’re here, we fulfill the promise of Remark 2.
Corollary 11. Fix a finite group \(G\). Then any representation of \(G\) is isomorphic to a direct sum of irreducible representations of \(G\).
Proof. By Theorem 1, it is enough to show that any representation is a direct sum of finite-dimensional representations. We will show this with transfinite induction. Fix some representation \(V\). The idea is to filter \(V\) by representations with finite-dimensional quotients.
To this end, let \(\{v_\alpha\}_{\alpha<\Lambda}\) be a basis of \(V\), where \(\Lambda\) is an ordinal with cardinality equal to the dimension of \(V\). Then for each \(\beta\le\Lambda\), let \(V_\beta\) be the subrepresentation of \(V\) generated by the basis vectors \(\{v_\alpha\}_{\alpha<\beta}\). Here are the properties of this filtration that we will need.
Now, we will show (by transfinite induction) that each \(V_\alpha\) is a direct sum of finite-dimensional representations, and the inclusion \(V_\beta\subseteq V_\alpha\) (for \(\beta<\alpha\)) is a surjection onto some of the components of \(V_\alpha\). The start of our indution is \(\alpha=0\), where \(V_\alpha=0\), so the claim has no content. We now have two inductive cases.
]]>Remark 12. The proof of Corollary 11 also shows that all irreducible representations are finite-dimensional.
In this post, all algebras are associative and unital, but they need not be commutative. Here is the classical statement of Schur’s lemma.
Proposition 1 (Schur’s lemma). Fix an algebra \(A\) over a field \(k\), and let \(M\) be an irreducible \(A\)-module.
- Then \(\operatorname{End}_A(M)\) is a division algebra.
- If \(M\) is finite-dimensional, then each \(T\in\operatorname{End}_A(M)\) is algebraic over \(k\).
Proof. We quickly explain how the second part follows from the first. Given that \(M\) is finite-dimensional over \(k\), it follows that \(\operatorname{End}_A(M)\) is a finite-dimensional division algebra over \(k\). Thus, for each \(T\in\operatorname{End}_A(M)\), we see that \(k(T)\subseteq\operatorname{End}_A(M)\) is some finite-dimensional field extension of \(k\), so \(T\) is algebraic over \(k\).
It remains to show the first part, for which we should show that any nonzero endomorphism \(T\) of \(M\) is invertible. It is enough to show that \(T\) is invertible as a \(k\)-linear map because then the inverse is automatically \(A\)-equivariant. Now, to show that \(T\) is invertible, it is enough to show that it has trivial kernel and cokernel. But this follows from irreducibility: \(\ker T\) and \(\operatorname{im}T\) are both \(A\)-submodules of \(M\) and thus must equal all of \(M\) by the irreducibility of \(M\). \(\blacksquare\)
Remark 2. To relate the second part to the usual formulation of Schur’s lemma, note that if \(k\) is algebraically closed, then \(T\in k\) is forced. It then follows that \(\operatorname{End}_A(M)=k\).
This post will be interested in when the second conclusion holds without assuming that \(M\) is finite-dimensional. This setting is frequently of interest when studying the representation theory of non-compact topological groups (such as non-compact Lie groups or \(p\)-adic groups). Here is the standard example explaining why some size condition may be necessary.
Example 3. Consider the algebra \(A=\mathbb C(x)\) over the field \(k=\mathbb C\). Then the module \(M=A\) is irreducible over \(A\) because \(A\) is a field, but \[\operatorname{End}_A(M)=\mathbb C(x).\]
Remark 4. Here is another indication that size plays a role: if \(M\) is small, then \(\operatorname{End}_A(M)\) automatically becomes smaller and hence closer to being \(k\). More precisely, we claim that \[\dim_k\operatorname{End}_A(M)\stackrel?\le\dim_kM.\] To see this, choose any nonzero vector \(v\in M\), and we claim that the map \(\operatorname{ev}_v\colon\operatorname{End}_A(M)\to M\) given by \(\operatorname{ev}_v\colon T\mapsto Tv\) is injective. Indeed, if \(\operatorname{ev}_vT=0\), then \(Tv=0\), so \(\ker T\subseteq M\) is a nonzero submodule, so \(\ker T=M\) because \(M\) is irreducible, so \(T=0\).
The problem in Example 3 turns out to be that \(\mathbb C(x)\) is a very large vector space.
Lemma 5. Fix a field \(k\). Then \[\dim_kk(x)=\left|k\right|+\aleph_0.\]
Proof. Note that \(\dim_kk(x)\le\left|k\right|+\aleph_0\) because \(k(x)\) has cardinality at most \(\left|k\right|+\aleph_0\). Indeed, by decomposing rational polynomials into numerators and denominators, the cardinality of \(k(x)\) is at most \(k[x]\times k[x]\), and the cardinality of \(k[x]\) is simply \(\left|k\right|+\aleph_0\) because \(k[x]\) can be identified with the countable union of finite products of \(k\).
It remains to show the other inequality, for which it is enough to check that \(\left|k\right|\le\dim_kk(x)\) and \(\aleph_0\le\dim_kk(x)\). Certainly \(\aleph_0\le\dim_kk(x)\) because \(k(x)\) contains the linearly independent elements \(\left\{1,x,x^2,\ldots\right\}\). For the other inequality, it is enough to check that the elements \[\left\{\frac1{x-c}:c\in k\right\}\] are linearly independent. Well, any relation \[\sum_{i=1}^n\frac{a_i}{x-c_i}=0\] can have its denominator cleared to show that \(x\) is the root of some nonzero polynomial in \(k[x]\), which is a contradiction. \(\blacksquare\)
We are now ready to state the usual generalization of Schur’s lemma, which is Dixmier’s lemma.
Proposition 6 (Dixmier’s lemma). Fix an algebra \(A\) over a field \(k\), and let \(M\) be an irreducible \(A\)-module. Suppose that \(\dim_kM<\left|k\right|\). Then any \(T\in\operatorname{End}_A(M)\) is algebraic over \(k\).
Proof. We proceed by contraposition. Suppose that we have an \(A\)-endomorphism \(T\) of \(M\) which is transcendental over \(k\); we will show that \(\dim_kM\ge\left|k\right|\). By Remark 4 and Lemma 5, it is enough to show that \[\dim_kk(x)\stackrel?\le\dim_k\operatorname{End}_A(M),\] for which it is enough to exhibit an embedding \(k(x)\to\operatorname{End}_A(M)\). We claim that we may define such a ring map by sending \(x\mapsto T\), which is then automatically injective because \(k(x)\) is a field. Now, to show that the canonical ring map \(k[x]\to\operatorname{End}_A(M)\) given by \(x\mapsto T\) extends to \(k(x)\), we have to show that the map is injective on \(k[x]\). This follows because \(T\) is transcendental over \(k\). \(\blacksquare\)
Remark 7. Here is a convenient way to check the dimension condition in Proposition 6. We claim that any irreducible \(A\)-module \(M\) has \(\dim_kM\le\dim_kA\). Indeed, it is enough to exhibit a surjection \(A\to M\), which we define as follows: fix any nonzero vector \(v\in M\), and then we can define \(A\to M\) by \(a\mapsto av\). To see that this map is surjective, we note that its image is an \(A\)-submodule of \(M\) and therefore must be all of \(M\).
Here is the usual application of Proposition 6.
Example 8. Fix a finite-dimensional Lie algebra \(\mathfrak g\) over \(\mathbb C\). Then \(U\mathfrak g\) is an algebra with countable dimension, so Remark 7 implies that any irreducible representation has at most countable dimension. We conclude from Proposition 6 (and Remark 2) that \[\operatorname{End}_{\mathfrak g}(M)=\mathbb C\] for any irreducible representation \(M\) of \(\mathfrak g\).
Of course, Proposition 6 has nothing to say about Lie algebras over \(\overline{\mathbb Q}\), which is somewhat surprising because one would expect that the representation theory over \(\overline{\mathbb Q}\) and \(\mathbb C\) would be basically the same. Descending to \(\overline{\mathbb Q}\) will require a bit more work.
Remark 9. Here is an approach which does not work: fix \(k=\overline{\mathbb Q}\), and choose an algebra \(A\) over \(k\). Then one may try to prove \(\operatorname{End}_A(M)=k\) for all irreducible \(A\)-modules \(M\) of at most countable dimension by base-changing along an embedding \(\overline{\mathbb Q}\to\mathbb C\) and then applying Proposition 6. But this cannot work because the claim that \(\operatorname{End}_A(M)=k\) is not always true: one can take \(A=M=k(x)\), as in Example 3. The reason why this approach fails is because \(M_{\mathbb C}\) need not be irreducible (and is not irreducible in the given counterexample).
Instead, we will follow Quillen. In contrast to Proposition 6, we will gain some control on sizes via a filtration.
Definition 10. An algebra \(A\) over a field \(k\) is filtered if and only if it admits an exhaustive, multiplicative, increasing filtration \(\{F^iA\}_{i\in\mathbb N}\) for which \(k\subseteq F^0A\).
Proposition 11 (Quillen’s lemma). Fix a filtered algebra \(A\) over a field \(k\). Suppose that \(\operatorname{gr}A\) is finitely generated and commutative over \(k\). Then any \(T\in\operatorname{End}_A(M)\) is algebraic over \(k\).
Example 12. We extend Example 8 to \(\overline{\mathbb Q}\). Fix a finite-dimensional Lie algebra \(\mathfrak g\) over \(\overline{\mathbb Q}\). By the Poincaré–Birkoff–Witt theorem, the natural degree filtration on \(U\mathfrak g\) makes \(\operatorname{gr}U\mathfrak g\) into a polyomial ring generated by a basis of \(\mathfrak g\). It follows from Proposition 11 (and Remark 2) that \[\operatorname{End}_{\mathfrak g}(M)=\overline{\mathbb Q}\] for any irreducible representation \(M\) of \(\mathfrak g\).
The proof of Proposition 11 is based on the following result from commutative algebra.
Theorem 13 (Generic freeness). Fix a Noetherian domain \(R\), and let \(S\) be an algebra of finite type over \(R\). For any finitely generated \(S\)-module \(M\), there is a nonzero \(f\in R\) such that \(M_f\) is free over \(R_f\).
Proof. The idea is to use Noether normalization to reduce to the case where \(S\) is a polynomial ring over \(R\) and \(M=S\). Quickly, we remark that we may localize all objects in sight by a nonzero element of \(R\) at any point and not adjust the validity of the conclusion.
Set \(K\) to be the fraction field of \(R\). We will proceed by induction on the Krull dimension \(d\) of \(S_K\), which we note is finite by Noether normalization (indeed, \(S_K\) is of finite type over \(K\)). For the base case, we note that if \(S_K=0\), then because \(S\) is finite type over \(R\), there is a nonzero \(f\in R\) for which \(S_f=0\), so it follows that \(M_f=0\), as required.
We reduce to the case where \(S\) is a polynomial ring over \(R\). We now apply Noether normalization, which grants elements \(x_1,\ldots,x_d\) of \(S_K\) such that \(K[x_1,\ldots,x_d]\subseteq S_K\) is a polynomial ring, and \(S_K\) is finite over \(K[x_1,\ldots,x_d]\). By clearing denominators in \(S_K\), we may assume that the \(x_\bullet\)s live in \(S\).
We now hunt for some nonzero \(c\in R\) so that \(S_c\) is finite over \(R_c[x_1,\ldots,x_d]\). Then \(M_c\) will continue to be finitely generated over \(R_c[x_1,\ldots,x_d]\), meaning that we have indeed reduced to the case where \(S_c\) is a polynomial ring over \(R_c\). To find this \(c\), we write \(S=R[s_1,\ldots,s_n]\) for some elements \(s_\bullet\in S\). Each \(s_i\) is integral over \(K[x_1,\ldots,x_d]\) and therefore satisfies a monic polynomial with coefficients in \(K[x_1,\ldots,x_d]\). Clearing denominators, we see that \(s_i\) satisfies some polynomial with coefficients in \(R[x_1,\ldots,x_d]\) whose leading coefficient is in \(R\). Then let \(c\) be the product of all the leading coefficients (to a sufficiently high power), and we find that \(s_i\) will be integral over \(R_c[x_1,\ldots,x_d]\). It follows that \(S_c\) is finite type and integral over \(R_c[x_1,\ldots,x_d]\) and therefore finite.
We reduce to the case where \(M\) takes the form \(S/\mathfrak q\) for a prime \(\mathfrak q\) of \(S\). By the theory of associated primes, \(M\) admits a filtration \[0=M_0\subseteq M_1\subseteq M_2\subseteq\cdots\subseteq M_m=M\] for which \(M_i/M_{i+1}\cong S/\mathfrak q_i\) for some prime \(\mathfrak q_i\) of \(S\) for each \(i\). Because an extension by a free module is free (recall that free modules are projective), and localization is exact, it is enough to establish the claim for each quotient \(M_i/M_{i+1}\), so we have reduced to the case where \(M\) takes the form \(S/\mathfrak q\) for a prime \(\mathfrak q\).
We complete the proof, using the reductions of the previous steps. There are now two cases.
Proof of Proposition 11. We may as well take \(T\ne0\). Because \(\operatorname{End}_A(M)\) is a division algebra (by Proposition 1), we see that \(k[T]\subseteq\operatorname{End}_A(M)\) is an integral domain. Eventually, we will show that \(k[T]\) is in fact a field, which we now note will complete the proof: then \(1/T\in k[T]\), so there is a polynomial \(p(x)\in k[x]\) for which \(Tp(T)-1=0\), which implies that \(T\) is algebraic.
We now proceed in steps.
We apply generic freeness. For any nonzero \(v\in M\), irreducibility implies that \(Av=M\). It follows that setting \[F^iM=k[T]\cdot F^iA\cdot v\] makes \(M\) into a finitely generated filtered module over \(k[T]\otimes_kA\). It follows that \(\operatorname{gr}M\) is finitely generated as a module over \(k[T]\otimes_k\operatorname{gr}A\) (by construction of the filtration), so Theorem 13 yields a nonzero \(f\in k[T]\) for which \((\operatorname{gr}M)_f\) is free over \(k[T]_f\).
We convert the freeness of \((\operatorname{gr}M)_f\) into a statement about \(k[T]\). Because localization is exact, we see that \((\operatorname{gr}M)_f=\operatorname{gr}M_f\), where \(M_f\) has been given the obvious (localized) filtration. By choosing generators of \(\operatorname{gr}M_f\) and arguing as in Lemma 8 of this post, we find that \(M_f\) is free over \(k[T]_f\).
Now, because \(\operatorname{End}_A(M)\) is a division algebra, we find that the fraction field \(k(T)\) naturally acts on \(M\), so \(M_f=M\). In fact, because \(k(T)\) is a field, we see that \(M\) is a direct sum of copies of \(k(T)\) as a \(k(T)\)-module and hence as a \(k[T]_f\)-module. It follows that \(k(T)\) is a direct summand of a free \(k[T]_f\)-module, so \(k(T)\) is free as a \(k[T]_f\)-module because \(k[T]_f\) is a principal ideal domain. (Here, \(k[T]_f\) is a principal ideal domain because it is a localization of a principal ideal domain.)
We now complete the proof. Thus far, we have shown that \(k(T)\) is free as a \(k[T]_f\)-module, and we want to show that \(k[T]\) is a field. Well, any two nonzero elements \(p_1(T)/q_1(T)\) and \(p_2(T)/q_2(T)\) of \(k(T)\) will admit a relation over \(k[T]_f\), so it follows that \(k(T)\) is free over \(k[T]_f\) of rank \(1\). Thus, the natural ring map \[k[T]_f\cong k(T)\] is an isomorphism. It follows that \(\dim k[T]_f=0\), so \(\dim k[T]=0\), so \(k[T]\) is a zero-dimensional principal ideal domain, so \(k[T]\) is a field. \(\blacksquare\)
I would guess that this argument is well-known, but I had a little trouble extracting it directly online, so I figured it would not be an injustice to write it down.
Proposition 1. Let \(S\) be a finite set of distinct primes. Then \[\operatorname{Gal}\left(\mathbb Q(\{\sqrt p:p\in S\})/\mathbb Q\right)\cong(\mathbb Z/2\mathbb Q)^{\#S}.\]
Proof. Our goal is to turn this Galois theory problem into group cohomology.
Let’s set up some notation. Set \(K=\mathbb Q(\{\sqrt p:p\in S\})\) and \(G=\operatorname{Gal}(\overline Q/\mathbb Q)\) for brevity. For each prime \(p\), let \(\chi_p\colon G\to\{\pm1\}\) denote the quadratic character given by the composite \[G=\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)\twoheadrightarrow\operatorname{Gal}(\mathbb Q(\sqrt p)/\mathbb Q)\cong\{\pm1\}.\] It will also be useful to define \(\chi\colon G\to\{\pm1\}^{\#S}\) by \(\chi=(\chi_p)_{p\in S}\).
We now proceed in steps.
We turn the theorem into a group theory problem. Here is the main point: by intersecting Galois subgroups via the Galois correspondence, we note that \(K\) is exactly the fixed field of \[\bigcap_{p\in S}\ker\chi_p\subseteq G.\] As such, \(\operatorname{Gal}(\overline Q/K)\) equals \(\ker\chi\). Thus, \(\chi\) descends to an injection \[\chi\colon\operatorname{Gal}(K/\mathbb Q)\to\{\pm1\}^{\#S}.\] To prove the theorem, it is enough to show that \(\chi\) is a bijection, which is a group theory problem.
We turn the theorem into a linear algebra problem: we claim that it is enough to check that the characters \(\{\chi_p\}_{p\in S}\) are linearly independent in the \(\mathbb F_2\)-vector space \(\operatorname{Hom}_{\mathrm{cts}}(G,\{\pm1\})\).
We proceed by contraposition. If \(\chi\) fails to be bijective, then it fails to be surjective, so its image lives in some hyperplane of \(\{\pm1\}^{\#S}\). Such a hyperplane is cut out by an equation, which yields some equation \[\prod_{p\in S}\chi_p^{a_p}=1,\] where each \(a_p\) is in \(\mathbb F_2\). This equation is exactly the failure of linear independence.
We do a little group cohomology. We are interested in the position of some elements in the cohomology group \[\operatorname{Hom}_{\mathrm{cts}}(G,\{\pm1\})=\mathrm H^1_{\mathrm{cts}}(G;\{\pm1\}).\] As such, we take Galois cohomology of the Kummer short exact sequence \[1\to\{\pm1\}\to\overline{\mathbb Q}^\times\stackrel 2\to\overline{\mathbb Q}^\times\to1\] to show that the boundary morphism \[\delta\colon\mathbb Q^\times\to\mathrm H^1_{\mathrm{cts}}(G;\{\pm1\})\] is surjective. (We have silently used Hilber’t theorem 90, which asserts \(\mathrm H^1_{\mathrm{cts}}(G;\overline{\mathbb Q}^\times)=0\).) Thus, there is an isomorphism \[\mathbb Q^\times/\mathbb Q^{\times2}\cong\mathrm H^1_{\mathrm{cts}}(G;\{\pm1\}),\] so we will be able to pass to the more controlled \(\mathbb F_2\)-vector space \(\mathbb Q^\times/\mathbb Q^{\times2}\).
We complete the proof. The main point is to show that \(\delta(p)=\chi_p\). Indeed, the boundary map \(\delta\) sends \(p\) to the \(1\)-cocycle \[\delta(p)\colon\sigma\mapsto\frac{\sigma(\sqrt p)}{\sqrt p},\] which is exactly \(\chi_p\): our \(\sigma\) goes to \(1\) if and only if it fixes \(\sqrt p\).
To complete the proof, we see that the previous step now allows us to merely prove that the elements \(\{p\}_{p\in S}\) are linearly independent in \(\mathbb Q^\times/\mathbb Q^{\times2}\), which is automatic from unique prime factorization in \(\mathbb Q\). \(\blacksquare\)
Remark 2. The proposition admits a straightforward generalization where primes are replaced by any set of pairwise coprime integers. Essentially the same proof works.
Remark 3. Here is an alternative, sillier way to solve the linear algebra problem. Note that \[\operatorname{Hom}_{\mathrm{cts}}(G;\{\pm1\})=\operatorname{Hom}_{\mathrm{cts}}(G^{\mathrm{ab}};\{\pm1\}).\] By the Kronecker–Weber theorem, this is isomorphic to \[\operatorname{Hom}_{\mathrm{cts}}(\widehat{\mathbb Z}^\times,\{\pm1\}).\] We can now show linear independence in this latter vector space. In this post, we showed that the Galois character \(\chi_p\) corresponds to a quadratic Dirichlet character \((\mathbb Z/4p\mathbb Z)^\times\to\{\pm1\}\). In particular, away from \(2\), the \(\chi_p\)s are supported on different factors of \[\operatorname{Hom}_{\mathrm{cts}}(\widehat{\mathbb Z}^\times,\{\pm1\})=\prod_p\operatorname{Hom}_{\mathrm{cts}}(\mathbb Z_p^\times,\{\pm1\}),\] so linear independence follows.
]]>Remark 4. The given proof of Theorem 1 is essentially an application of Kummer theory, so it will generalize well to handle cyclic (or even abelian) extensions with enough roots of unity in the base field. In contrast, Remark 3 is essentially an application of class field theory, so it could conceivably generalize to handle arbitrary abelian extensions.
Our exposition very loosely follows Anna Cadoret.
Theorem 1. Fix an abelian variety \(A\) over an algebraically closed field of characteristic zero. Then \[\pi_1^{\mathrm{\acute et}}(A;0)=\lim_n A[n].\]
We will not review the entire theory of the étale fundamental group, but we will recall its basic properties. The general theory is rather abstract and bland, so we will content ourselves with simply stating what we need. For a quick exposition, see (for example) here.
Definition 2. Fix a scheme \(X\), and choose a geometric point \(\overline x\hookrightarrow X\). Let \(\mathrm{F\acute Et}(X)\) denote the category of finite étale covers of \(X\). Then the étale fundamental group \(\pi_1^{\mathrm{\acute et}}(X;\overline x)\) is the automorphism group of the fiber functor \[(-)_{\overline x}\colon\mathrm{F\acute Et}(X)\to\mathrm{FinSet}\] given by sending a finite étale cover \(Y\to X\) to the fiber \(Y_{\overline x}\to\overline x\).
Remark 3. It turns out that the functor \((-)_{\overline x}\) upgrades into an equivalence of categories \[(-)_{\overline x}\colon\mathrm{F\acute Et}(X)\to\mathrm{FinSet}\left(\pi_1^{\mathrm{\acute et}}(X;\overline x)\right).\]
Example 4. Consider \(X=\operatorname{Spec}k\), where \(k\) is some field. Then the category of finite étale covers of \(X\) is the opposite category of the category of finite-dimensional separable \(k\)-algebras. By taking embeddings into a fixed algebraic closure (which corresponds to a chosen geometric point of \(X\)), one can check that \(\pi_1^{\mathrm{\acute et}}(X;\overline x)=\operatorname{Gal}(k^{\mathrm{sep}}/k)\).
This is a great definition, but it makes it very unclear how one might compute anything. To this end, it is enough to compute the automorphisms of \(F\) on a subcategory of \(\mathrm{F\acute Et}(X)\). In particular, it turns out to be enough to work with a cofinal sequence of Galois objects.
Definition 5. A finite étale cover \(Y\to X\) is Galois if and only if its automorphism group equals its degree.
Example 6. A Galois extension of fields produces a Galois cover.
Example 7. Let \(A\) be an abelian variety over a field \(k\). If an integer \(n\) is coprime to the characteristic of \(n\), then the multiplication-by-\(n\) map \([n]\colon A\to A\) is finite and smooth and thus a finite étale cover. It turns out to have degree \(n^{\dim A}\) with automorphism group isomorphic to its kernel (acting by translation), which is \[A[n]\cong(\mathbb Z/n\mathbb Z)^{2\dim A}.\] Thus, \([n]\colon A\to A\) is a Galois cover with automorphism group \(A[n]\).
Remark 8. Let \(\mathcal C\) be a cofinal subcategory of \(\mathrm{F\acute Et}(X)\) of Galois objects, where cofinal means that any object in \(\mathrm{F\acute Et}(X)\) admits a morphism from an object in \(\mathcal C\). It follows from the general theory that the natural map \[\pi_1^{\mathrm{\acute et}}(X;\overline x)\to\lim_{Y\in\mathcal C}\operatorname{Aut}_XY\] is an isomorphism.
The main input to Theorem 1 will be a classification of finite étale covers of abelian varieties.
Lemma 9. Let \(A\) be an abelian variety over an algebraically closed field \(k\). Then \(\pi_1^{\mathrm{\acute et}}(A;0)\) is abelian.
Proof. By a variant of the Eckmann–Hilton argument (see this blog post), it is enough to show that \(\pi_1^{\mathrm{\acute et}}(A;0)\) is a monoid object (with identity). (This implies that \(\pi_1^{\mathrm{\acute et}}(A;0)\) is a monoid object in the category of groups, which forces the group to be abelian.) Well, functoriality provides both a natural identity via \[\pi_1^{\mathrm{\acute et}}(\operatorname{Spec}k)\stackrel0\to\pi_1^{\mathrm{\acute et}}(A;0)\] and multiplication via \[\pi_1^{\mathrm{\acute et}}(A)\times\pi_1^{\mathrm{\acute et}}(A)\stackrel{(i_1,i_2)}\to\pi_1^{\mathrm{\acute et}}(A\times A)\stackrel m\to\pi_1^{\mathrm{\acute et}}(A).\] (All basepoints are the zero point.) All these maps are group homomorphisms, and functoriality can verify that the identity map is in fact an identity using the group law on \(A\). The claim follows. \(\blacksquare\)
Proposition 10. Let \(A\) be an abelian variety over an algebraically closed field. For any connected finite étale cover \(Y\to A\) and point \(\overline y\in Y_0\), the scheme \(Y\) can be given the structure of an abelian variety for which \(Y\to A\) is an isogeny.
Proof. The main point is to show that \(Y\) is an abelian variety. Note that \(Y\) is connected, \(Y\) is reduced because \(A\) is reduced, and \(Y\) is proper because \(Y\to A\) is finite. Thus, it is enough to show that \(Y\) is a group scheme and that \(Y\to A\) is a morphism of group schemes.
By the equivalence of categories in Remark 3, it is enough to exhibit a group object structure on \(Y_0\). Indeed, once this is achieved, we also automatially know that \(Y_0\to A_0\) is a morphism of group objects (because the target is the trivial group), so Remark 3 informs us that we have also constructed a morphism of group objects in the category \(\mathrm{F\acute Et}(A)\).
Now, to put a group object structure on \(Y_0\), we note that \(Y\) is a connected object, so \(\pi_1^{\mathrm{\acute et}}(A;0)\) acts transitively on \(Y_0\). Thus, there is a (closed) finite-index subgroup \(H\subseteq\pi_1^{\mathrm{\acute et}}(A;0)\) (equal to the stabilizer of \(\overline y\)) for which there is an isomorphism \[\pi_1^{\mathrm{\acute et}}(A;0)/H\to Y_0\] of \(\pi_1^{\mathrm{\acute et}}(A;0)\)-sets given by \(g\mapsto g\overline y\). Because \(\pi_1^{\mathrm{\acute et}}(A;0)\) is an abelian group (by Lemma 9), \(H\) is normal, so the left-hand side can be given a natural group quotient group structure. Thus, \(Y_0\) is equipped with a natural quotient group structure as well (which is notably \(\pi_1^{\mathrm{\acute et}}(A;0)\)-equivariant), and we are done! \(\blacksquare\)
We are now ready to attack Theorem 1 directly, via Remark 8.
Corollary 11. Let \(A\) be an abelian variety over an algebraically closed field of characteristic zero. The finite étale covers \([n]\colon A\to A\) form a cofinal sequence.
Proof. For any finite étale cover \(\varphi\colon Y\to A\), Proposition 10 informs us that \(Y\) can automatically be upgraded into an abelian variety, and \(\varphi\colon Y\to A\) can automatically be upgraded into an isogeny.
Now, say that \(\varphi\) has degree \(n\). Then \(\ker\varphi\) is annihilated by \([n]\), so because \(\varphi\colon Y\to A\) can be viewed as an fppf quotient (by \(\ker\varphi\subseteq Y\) acting by translation), it follows that \([n]\colon A\to A\) factors through \(\varphi\). In other words, there is an isogney \(\psi\colon A\to Y\) so that the composite \[A\to Y\to A\] is \([n]\). It follows that \(Y\to A\) is covered by \([n]\colon A\to A\). \(\blacksquare\)
Proof of Theorem 1. By Remark 8 and Corollary 11, we see that \[\pi_1^{\mathrm{\acute et}}(A;0)=\lim_n\operatorname{Aut}([n]\colon A\to A).\] This latter automorphism group has been computed to be \(A[n]\), acting by translation, in Example 7. Before calling it quits, we remark that the transition maps are as follows: \([nm]\colon A\to A\) covers \([n]\colon A\to A\) via the map \([m]\colon A\to A\); on the Galois groups, this corresponds to the covering \([m]\colon A[mn]\to A[n]\) by examining the “restriction” of the translation action of \(A[mn]\) on \(A\) along the cover \([m]\colon A\to A\). \(\blacksquare\)
]]>Remark 12. Let’s explain how to relax the hypothesis on \(\operatorname{char}k\). The same argument shows that the \(\ell\)-primary part of \(\pi_1^{\mathrm{\acute et}}(A;0)\) is isomorphic to \[T_\ell A=\lim A[\ell^\bullet]\] as long as \(\ell\) is coprime to \(\operatorname{char}k\). This equality in fact turns out to be true when \(\ell=\operatorname{char}k\), but it is more subtle to prove because \([\ell]\colon A\to A\) is no longer smooth: instead, one needs to factor out the Frobenius enough times to produce the needed cofinal sequence of finite étale covers.
Our exposition largely follows Andrew Sutherland. We take the following (albeit nonstandard) definition.
Definition 1. Fix number fields \(K\) and \(K’\). Then \(K\) and \(K’\) are arithmetically equivalent if and only if, for all but finitely many rational primes \(p\), the factorization types of \(p\mathcal O_K\) and \(p\mathcal O_{K’}\) are the same. We may write this as \(K\sim K’\).
Our main goal will be to present the following permanence properties.
Theorem 2. The following quantities are preserved by arithmetic equivalence.
- The Dedekind \(\zeta\)-function.
- The set of unramified primes and the factorization types for all unramified primes.
- The number of real embeddings and the number of complex embeddings.
- The degree.
- The discriminant.
- The normal closure.
- The normal core, which is the largest subfield Galois over \(\mathbb Q\).
- The isomorphism class of the group of units.
The most difficult of the equivalences in Theorem 2 will be the Dedekind \(\zeta\)-function.
Remark 3. For a sense of scale, we note that \(K\sim K’\) already implies that \(\zeta_K\) and \(\zeta_{K’}\) are the same up to finitely many Euler factors. Indeed, if \(p\) is a rational prime, then the Euler factor of \(\zeta_K\) above \(p\) is given by \[\prod_{\mathfrak p\mid p\mathcal O_K}\frac1{1-\operatorname N(\mathfrak p)^{-s}},\] so this Euler factor is entirely determined by the number of primes \(p\) splits into and the inertia degrees of those primes.
The main input to Theorem 2 will be the functional equation.
Definition 4. Fix a number field \(K\) of signature \((n_1,n_2)\). Then the completed Dedekind \(\zeta\)-function is \[\Xi_K(s)=\left|\operatorname{disc}\mathcal O_K\right|^{s/2}\Gamma_{\mathbb R}(s)^{n_1}\Gamma_{\mathbb C}(s)^{n_2}\zeta_K(s),\] where \(\Gamma_{\mathbb R}(s)=\pi^{-s/2}\Gamma(s/2)\) and \(\Gamma_{\mathbb C}(s)=2(2\pi)^{-s}\Gamma(s)\).
Theorem 5. Fix a number field \(K\). Then \(\zeta_K(s)\) admits a meromorphic continuation to \(\mathbb C\), which is holomorphic everywhere except for a simple pole at \(s=1\). Furthermore, there is the functional equation \[\Xi_K(s)=\Xi_K(1-s).\]
Proof. This is a difficult theorem and thus omitted. \(\blacksquare\)
Proposition 6. Fix number fields \(K\) and \(K’\). Then the following are equivalent.
- \(K\) and \(K’\) are arithmetically equivalent.
- \(\zeta_K=\zeta_{K’}\).
Proof. As discussed in Remark 3, arithmetic equivalence is equivalent to having the equality of \(\zeta_K\) and \(\zeta_{K’}\) up to finitely many Euler factors. Namely, we are using the fact that the coefficients of a Dirichlet series can be recovered from the holomorphic function by integrating in vertical strips. Once the coefficients of the Dirichlet series are recovered, one can use the \(p\)-power coefficients to recover the Euler factor at \(p\) and hence the factorization type for unramified primes.
Thus, it only remains to show that having an equality up to finitely many Euler factors implies full equality. To this end, let \(S\) be a finite set of rational primes containing all rational primes where the Euler factors of \(\zeta_K\) and \(\zeta_{K’}\) disagree. Now, let \(D_K\) and \(D_{K’}\) be the absolute discriminants, and let \((n_1,n_2)\) and \((n_1’,n_2’)\) be the signatures of \(K\) and \(K’\) respectively. Then \[\frac{\Xi_K(s)}{\Xi_{K’}(s)}=\left|\frac{D_K}{D_{K’}}\right|^{s/2}\Gamma_{\mathbb R}(s)^{n_1-n_1’}\Gamma_{\mathbb C}(s)^{n_2-n_2’}\prod_{p\in S}\frac{E_{K’,p}(p^{-s})}{E_{K,p}(p^{-s})},\] where \(E_{K,p}(p^{-s})\) and \(E_{K’,p}(p^{-s})\) are the Euler factors. We now hit this equality with the functional equation and compare zeroes and poles.
The above factor-by-factor analysis reveals that \(\Xi_K(s)/\Xi_{K’}(s)\) and \(\Xi_K(1-s)/\Xi_{K’}(1-s)\) have disjoint sets of zeroes and poles. But these are of course the same function by Theorem 5, so we are forced to conclude that these quotients have no zeroes or poles or at all!
We will be able to convert the previous paragraph into the result after a little work. Note that \(\frac{2\pi i}{\log p}\cdot\frac ab\) will be a root of \(E_{K,p}(p^{-s})\) exactly with multiplicity equal to the number of inertial degrees \(f(\mathfrak p/(p))\) dividing \(b\) (where \(a/b\) is assumed to be in reduced form). Thus, any nontrivial factor in the product \(\prod_{p\in S}E_{K’,p}(p^{-s})/E_{K,p}(p^{-s})\) will contribute roots or poles with nonzero imaginary part. This is illegal, so we conclude that all terms must be \(1\). It follows that \(\zeta_K=\zeta_{K’}\). \(\blacksquare\)
The proof above actually equips us to show much more of Theorem 2.
Corollary 7. Fix number fields \(K\) and \(K’\). If \(K\sim K’\), then \(K\) and \(K’\) have the same signature and degree.
Proof. Because the degree is twice the number of total number of complex embeddings, it is enough to show the equality of signatures, which we do in two steps.
Remark 8. Here is a more direct proof of the degree equality: choose any rational prime \(p\) which is unramified in both \(K\) and \(K’\) and admits the same factorization type in both \(K\) and \(K’\). Then \(p\) splits into \(r\) primes, which we may say have inertia degrees \((f_1,\ldots,f_r)\). But the fundamental identity shows us that \[[K:\mathbb Q]=\sum_{i=1}^rf_i=[K’:\mathbb Q],\] as desired.
Corollary 9. Fix number fields \(K\) and \(K’\). If \(K\sim K’\), then \(K\) and \(K’\) have the same discriminant.
Proof. With an equality of signatures and Dedekind \(\zeta\)-functions, we see that \[\frac{\Xi_K(s)}{\Xi_{K’}(s)}=\left|\frac{\operatorname{disc}\mathcal O_K}{\operatorname{disc}\mathcal O_{K’}}\right|^s.\] The functional equation of Theorem 5 thus forces \(\left|\operatorname{disc}\mathcal O_K\right|=\left|\operatorname{disc}\mathcal O_{K’}\right|\).
It remains to pin down the equality of signs. We will show that the sign of the discriminant depends only on the number of complex embeddings, which will complete the proof by Corollary 7. Because the discriminant of an element of \(K\) will differ from \(\operatorname{disc}\mathcal O_K\) only up to a square, it is enough to compute the sign of \(\operatorname{disc}\alpha\) where \(K=\mathbb Q(\alpha)\). For this, let \(\{\sigma_1,\ldots,\sigma_n\}\) be the set of embeddings, so \[\operatorname{disc}\alpha=\det\begin{bmatrix}\sigma_1(1) & \cdots & \sigma_1(\alpha^{n-1}) \\\vdots & \ddots & \vdots \\\sigma_n(1) & \cdots & \sigma_n(\alpha^{n-1})\end{bmatrix}^2,\] where \(n\) is the degree of \(K\).
Now, consider the action by complex conjugation on the determinant \[\det\begin{bmatrix}\sigma_1(1) & \cdots & \sigma_1(\alpha^{n-1}) \\\vdots & \ddots & \vdots \\\sigma_n(1) & \cdots & \sigma_n(\alpha^{n-1})\end{bmatrix},\] which is the square root of the discrimimant. On one hand, complex conjugation will fix the square root if and only if \(\operatorname{disc}\alpha>0\). On the other hand, compex conjugation will swap conjugate rows of complex embeddings while fixing rows of real embeddings, in effect multiplying the entire determinant by \((-1)^{n_2}\), where \(n_2\) is the number of pairs of complex embeddings. We conclude that \[\operatorname{sgn}\operatorname{disc}\mathcal O_K=(-1)^{n_2},\] as required. \(\blacksquare\)
Corollary 10. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then the normal closures of \(K\) and \(K’\) are isomorphic.
Proof. Let \(L\) and \(L’\) be the normal closures of \(K\) and \(K’\), respectively.
We will use the fact that totally split primes recognize Galois extensions. To be explicit, let \(\operatorname{Spl}(M)\) denote the set of rational primes which are totally split in a number field \(M\). Then being totally split is preserved by taking the normal closure, so \(\operatorname{Spl}(K)=\operatorname{Spl}(L)\) and \(\operatorname{Spl}(K’)=\operatorname{Spl}(L’)\). But now recall from this post that having the equality \[\operatorname{Spl}(L)=\operatorname{Spl}(L’)\] shows that \(L\cong L’\), so we are done because checking Euler factors of Proposition 6 yields this equality. \(\blacksquare\)
Remark 11. The above proof is dishonest because one does not actually need to appeal to Proposition 6. Indeed, to conclude that \(L=L’\), one technically only needs the equality \[\operatorname{Spl}(L)=\operatorname{Spl}(L’)\] up to a finite number of primes.
In this subsection, we will provide a group-theoretic characterization of arithmetic equivalence, and it will help us prove the rest of Theorem 2.
To start, we provide a group-theoretic way to compute prime-splitting in non-Galois extensions using Frobenius elements.
Proposition 12. Let \(K\) be a number field, and let \(L\) be an extension of \(K\) which is Galois over \(\mathbb Q\). Set \(G=\operatorname{Gal}(L/\mathbb Q)\) and \(H=\operatorname{Gal}(L/K)\) for brevity. Fix a rational prime \(p\) unramified in \(L\), and choose a prime \(\mathfrak q\) of \(L\) above it.
- The map \(G\to\{\mathfrak p\subseteq\mathcal O_K:\mathfrak p\mid p\mathcal O_K\}\) given by \(\sigma\mapsto(\sigma\mathfrak q\cap K)\) descends to a bijection from the double coset space \(H\backslash G/D_{\mathfrak q}\).
- The inertia degree of the prime \(\sigma\mathfrak q\cap K\) over \(p\) is the size of \(H\backslash H\sigma D_{\mathfrak q}\).
Proof. We show the two parts separately.
For the surjectivity of the map, we note that any prime \(\mathfrak p\) of \(K\) above \(p\) lies under some prime \(\mathfrak q’\) of \(L\). But \(G\) acts transitively on the primes of \(L\) above \(p\), so we may find \(\sigma\in G\) with \(\mathfrak q’=\sigma\mathfrak q\), whereupon \(\mathfrak p=\sigma\mathfrak q\cap K\) follows.
It remains to show injectivity. Well, suppose that \(\sigma\mathfrak q\cap K=\tau\mathfrak q\cap K\) for \(\sigma,\tau\in G\). Then there are two primes \(\sigma\mathfrak q\) and \(\tau\mathfrak q\) of \(L\) which lie over the same prime of \(K\), so there is an element \(h\in H\) for which \(h\sigma\mathfrak q=\tau\mathfrak q\). It follows that \(\tau^{-1}h\sigma\in D_{\mathfrak q}\), so \(H\sigma D_{\mathfrak q}=H\tau D_{\mathfrak q}\).
Everything is unramified, so \(D_{\mathfrak q}\) is cyclic (generated by the Frobenius), so the size \(f\) of \(H\backslash H\sigma D_{\mathfrak q}\) is exactly the least positive integer for which \[H\sigma\mathrm{Frob}_{\mathfrak q}^f=H\sigma.\] Rearranging, this is the least positive integer for which \(\mathrm{Frob}_{\sigma\mathrm q}^f\in H\). By reducing (which will not affect nontriviality because everything is unramified), we see that this is the least power of the Frobenius which acts trivial on the residue field \(\sigma\mathrm q\cap L^H\). This is exactly the inertia degree of \(\sigma\mathrm q\cap K\). \(\blacksquare\)
Here is our group-theoretic characterization.
Definition 13. Two subgroups \(H\) and \(H’\) of \(G\) are Gassmann equivalent if and only if one has an equality \[\{\left|H\backslash H\sigma C\right|:H\sigma C\in H\backslash G/C\}=\{\left|H’\backslash H’\sigma C\right|:H’\sigma C\in H’\backslash G/C\}\] of multisets for any cyclic subgroup \(C\subseteq G\).
Corollary 14. Fix number fields \(K\) and \(K’\). Let \(L\) be a Galois extension of \(\mathbb Q\) containing \(KK’\) with Galois group \(G\), and set \(H=\operatorname{Gal}(L/K)\) and \(H’=\operatorname{Gal}(L/K’)\). The following are equivalent.
- \(K\) and \(K’\) are arithmetically equivalent.
- \(H\) and \(H’\) are Gassmann equivalent.
Proof. By Proposition 12, the factorization type in \(K\) of any unramified rational prime \(p\) is given exactly by the multiset \[\{\left|H\backslash H\sigma\langle\mathrm{Frob}_{\mathfrak q}\rangle\right|:H\sigma C\in H\backslash G/\langle\mathrm{Frob}_{\mathfrak q}\rangle,\}\] where \(\mathfrak q\) is some chosen prime in \(L\) above \(p\). Thus, Gassmann equivalence implies having the same factorization type for all unramified primes, which implies arithmetic equivalence.
Conversely, arithmetic equivalence implies having equality \[\{\left|H\backslash H\sigma C\right|:H\sigma C\in H\backslash G/C\}=\{\left|H’\backslash H’\sigma C\right|:H’\sigma C\in H’\backslash G/C\}\] whenever \(C=\langle\mathrm{Frob}_{\mathfrak q}\rangle\). By the Chebotarev density theorem, any cyclic subgroup takes this form, so Gassmann equivalence follows. \(\blacksquare\)
Here is a more useful way to test Gassmann equivalence.
Proposition 15. Fix subgroups \(H\) and \(H’\) of a group \(G\). Then the following are equivalent.
- \(H\) and \(H’\) are Gassmann equivalent.
- For each \(\sigma\in G\), we have \[\left|\{H\tau:H\tau=H\tau\sigma\}\right|=\left|\{H’\tau:H’\tau=H’\tau\sigma\}\right|.\]
- There is an isomorphism \(\mathbb C[H\backslash G]\cong\mathbb C[H’\backslash G]\) of (permutation) representations of \(G\).
- For any conjugacy class \(c\), we have \(\left|c\cap H\right|=\left|c\cap H’\right|\).
Proof. We start with the equivalence of the first three points.
It remains to show equivalence with the fourth point. We will show that the second and fourth points are equivalent. Quickly, note that we always have \(\left|H\right|=\left|H’\right|\): the fourth point implies this by summing over conjugacy classes, and the third point implies this by taking dimensions.
Now, for any \(\sigma\in G\), let \(G\cdot\sigma\) denote the conjugacy class. To start, note \[\left|H\cap G\cdot\sigma\right|\cdot\left|C(\sigma)\right|=\left|\{\tau\in G:\tau\sigma\tau^{-1}\in H\}\right|\] by the Orbit–stabilizer theorem. The right-hand side can be rearranged into \[\left|\{\tau\in G:H\tau\sigma=H\tau\}\right|,\] which of course is \[\left|H\cdot\right|\cdot\left|\{H\tau\in H\backslash G:H\tau\sigma=H\tau\}\right|.\] We conclude that \(\left|H\cap G\cdot\sigma\right|\) and \(\left|\{H\tau\in H\backslash G:H\tau\sigma=H\tau\}\right|\) disagree only by some factor depending on \(\left|H\right|\) and \(\sigma\), so the equivalence of the second and fourth points follows. \(\blacksquare\)
Corollary 16. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then the normal cores of \(K\) and \(K’\) are isomorphic.
Proof. Let \(L\) be the Galois closure of \(KK’\). Then the normal sub-extensions of \(L/\mathbb Q\) are in bijection with the normal subgroups of \(G=\operatorname{Gal}(L/\mathbb Q)\), so the normal core of \(K\) is given by the fixed field of the smallest normal subgroup of \(G\) containing \(H=\operatorname{Gal}(L/K)\). An analogous statement holds for the normal core of \(K’\) with the subgroup \(H’=\operatorname{Gal}(L/K’)\). We are thus left to show that \[N_G(H)\stackrel?=N_G(H’).\] To see this, note the normalizer of a subgroup \(H\) is formed by (iteratively) taking the group generated by the conjugacy classes intersecting \(H\). Because the conjugacy classes intersecting \(H\) and \(H’\) are the same by combining Corollary 14 and Proposition 15, we conclude that the normalizers are the same, so we are done. \(\blacksquare\)
Corollary 17. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then \(\mathcal O_K^\times\cong\mathcal O_{K’}^\times\).
Proof. Note that the unit groups are finitely generated abelian groups, so their isomorphism type is given by their torsion groups and their ranks. The rank is given by Dirichlet’s unit theorem: it is determined by the signature, which is the same for \(K\) and \(K’\) by Corollary 7.
It remains to show that \(\mathcal O_{K,\mathrm{tors}}^\times\cong\mathcal O_{K’,\mathrm{tors}}^\times\). Well, the torsion subgroup of a unit group is exactly the roots of unity, so it is enough to check that \[K\cap\mathbb Q(\mu_\infty)\stackrel?=K’\cap\mathbb Q(\mu_\infty),\] where \(\mathbb Q(\mu_\infty)\) is the cyclotomic closure of \(\mathbb Q\). However, \(K\cap\mathbb Q(\mu_\infty)\) is abelian over \(\mathbb Q\) and will therefore be contained in the normal core of \(K\), so Corollary 16 tells us that \(K\cap\mathbb Q(\mu_\infty)\subseteq K’\). Thus, \[K\cap\mathbb Q(\mu_\infty)\subseteq K’\cap\mathbb Q(\mu_\infty),\] and the other inclusion follows by symmetry. \(\blacksquare\)
]]>Remark 18. Suppose \(K\sim K’\) so that \(\zeta_K=\zeta_{K’}\) by Proposition 6. Then the residues at \(s=1\) are the same, allowing us to compare class number formulae of \(K\) and \(K’\). Theorem 2 shows that many of the quantities present in the class number formula are already the same, leaving us only with \[\left|\mathrm{Cl}_K\right|\cdot\left|\mathrm{Reg}_K\right|=\left|\mathrm{Cl}_{K’}\right|\cdot\left|\mathrm{Reg}_{K’}\right|.\] However, it turns out that the class number alone is not determined by arithmetic equivalence class.
Theorem 1. Fix a prime \(p\) such that \(p\equiv1\pmod3\).
- There is a unique pair of integers \((A,B)\) for which \(4p=A^2+27B^2\) and \(A\equiv2\pmod3\) and \(B>0\).
- There is a unique cubic subfield of \(\mathbb Q(\zeta_p)\), and it is generated by a root of the equation \(x^3-3px+pA\).
Quickly, we remark that one can adjust the sign of \(A\) to instead have \(A\equiv1\pmod3\), which makes the equation at the end \(x^3-3px-pA\).
We will show the two parts more or less independently. Because it is easier, let’s start with the first part. Throughout today, \(\omega\) is a primitive third root of unity.
Lemma 2. Fix a nonzero element \(\alpha\) of \(\mathbb Z[\omega]\) which is coprime to \(3\). Then \(\alpha\) admits a unique associate \(\alpha’\) such that \(\alpha’\equiv1\pmod3\).
Proof. We are only interested in\(\pmod3\) information, so it is enough to show that the composite map \[\mu_6=\mathbb Z[\omega]^\times\to\left(\frac{\mathbb Z[\omega]}{(3)}\right)^\times\] is an isomorphism. (Indeed, given \(\alpha\), we are hunting for a unit \(u\) for which \(u\alpha\equiv1\pmod3\), which is equivalent to having a unique unit \(u\) with \(u\equiv\alpha^{-1}\pmod3\).) To show this, \(\mathbb Z[\omega]/(3)\) is represented by the elements \[\left\{\pm1,\pm\omega,\pm(1+\omega)\right\}\sqcup\{0,\pm(1-\omega)\}.\] The left set is \(\mu_6\), and the elements of the right set share a prime factor with \(3\) because \(3=(1-\omega)(2+\omega)\). \(\blacksquare\)
Proposition 3. Fix a prime \(p\) such that \(p\equiv1\pmod3\). There is a pair of integers \((A,B)\) for which \(4p=A^2+27B^2\) and \(A\equiv2\pmod3\).
Proof. Because \(p\equiv1\pmod3\), we see that \(\left(\frac{-3}p\right)=+1\), so \[\frac{\mathbb Z[\omega]}{(p)}\cong\frac{\mathbb F_p[x]}{\left(x^2+3\right)}\] is not a field, so \(p\) is not a prime element in the principal ideal domain \(\mathbb Z[\omega]\), so we may factor \(p=\pi\overline\pi\) for some \(\pi\). By Lemma 2, we may adjust \(\pi\) by some root of unity so that \(\pi\equiv1\pmod3\).
Now, write \(\pi=a-b\omega\) so that \(\operatorname N_{\mathbb Q(\omega)/\mathbb Q}(\pi)=a^2+ab+b^2\), so \[p=a^2+ab+b^2,\] which rearranges into \[4p=(2a+b)^2+3b^2.\] To finish, note \(\pi\equiv1\pmod3\) requires that \(a\equiv1\pmod3\) and \(b\equiv0\pmod3\), so we may take \((A,B)=(2a+b,b/3)\). \(\blacksquare\)
Remark 4. Conversely, if we write \(4p=A^2+27B^2\) as in Theorem 1, then we may set \(b=3B\) and \(a=(A-b)/2\); note that \(a\) is in fact an integer because \(A\) and \(B\) must have the same parity. Then \(p=a^2+ab+b^2\), so \[\pi=a-b\omega=\frac{A+3B\sqrt{-3}}2\] is a factor of \(p\) with \(\pi\equiv1\pmod3\). By unique prime factorization of \(p\), we see that this \(\pi\) is unique up to conjugation. It follows that the pair \((A,B)\) is unique up to the sign of \(B\).
We now turn to the calculation of the cubic subfield.
Proposition 5. Let \(L/K\) be a finite Galois extension of fields with Galois group \(G\), and choose a subgroup \(H\subseteq G\).
- For each \(\theta\), we have \[\sum_{h\in H}h\theta\in L^H.\]
- If \(\theta\in L\) is a normal basis element (i.e., \(\{g\theta:g\in G\}\) is a basis for \(L\) over \(K\)), then \[L^H=K\left(\sum_{h\in H}h\theta\right).\]
Proof. Note that \(g\in G\) fixes \(\sum_{h\in H}h\theta\) if and only if \[\sum_{h\in H}gh\theta=\sum_{h\in H}h\theta.\] For example, if \(g\in H\), then the sum is merely rearranged, so the first part follows. For the second part, having a basis means that the equality in \(L\) is equivalent to an equality \[\{gh\theta:h\in H\}=\{h\theta:h\in H\}\] of sets. In particular, \(g\theta\) is in the right-hand set, so \(g\in H\) follows; conversely, \(g\in H\) implies that the two sets are merely permutations of each other. \(\blacksquare\)
Corollary 6. Fix a prime \(p\equiv1\pmod3\). Then there is a unique cubic subfield of \(\mathbb Q(\zeta_p)\), and it is generated by \[\sum_{i=1}^{p-1}\zeta_p^{i^3}.\]
Proof. Let’s start with the existence and uniqueness of the cubic subfield. The Galois group \(G\) of the extension \(\mathbb Q(\zeta_p)/\mathbb Q\) is the cyclic group \((\mathbb Z/p\mathbb Z)^\times\). The uniqueness of the cubic subfield follows because cyclic groups have unique subgroups of given index (dividing the order). In particular, the unique subgroup of \((\mathbb Z/p\mathbb Z)^\times\) consists of the cubes \[H:=\left\{k^3:k\in(\mathbb Z/p\mathbb Z)^\times\right\}.\] It remains to show that the given element generates \(\mathbb Q(\zeta_p)^H\), which is the purpose of Proposition 5. Indeed, the element \(\zeta_p\) produces a basis \(\{\zeta^i:i\in(\mathbb Z/p\mathbb Z)^\times\}\) for the extension \(\mathbb Q(\zeta_p)/\mathbb Q\), so we see that \[3\sum_{h\in H}h\zeta_p\] generates \(\mathbb Q(\zeta_p)^H\). Because cubing on \((\mathbb Z/p\mathbb Z)^\times\) is \(3\)-to-\(1\) onto \(H\), the result follows. \(\blacksquare\)
It remains to compute the cubic irreducible polynomial of the element defined in Corollary 6. For this, we recall (e.g., from Fourier analysis or representation theory) that it is convenient to replace a sum over an indicator of a subgroup with some character sum. Namely, for a finite abelian group \(A\), the orthogonality relations \[\frac1{\left|A\right|}\sum_{\chi\in\operatorname{Hom}(A,\mathbb C)}\chi\] is the indicator of the identity of \(A\). (We will prove the needed special case of this later.) Because we want to indicate the subgroup of cubes in \((\mathbb Z/p\mathbb Z)^\times\), we want to understand characters of the quotient, which are equivalent to characters \((\mathbb Z/p\mathbb Z)^\times\to\mu_3\).
We will need to do some intricate calculations with our characters, so we take some care in defining our character. The following construction is motivated by Euler’s criterion.
Definition 7. Fix a prime \(\pi\) of \(\mathbb Z[\omega]\) with \(\pi\nmid3\). Then we define \(\chi_\pi\colon(\mathbb Z[\omega]/(\pi))^\times\to\mu_3\) by \[\alpha^{(\operatorname N(\pi)-1)/3}\equiv\chi_\pi(\alpha)\pmod\pi\] for each \(\alpha\). The fraction here is an integer: one can check that \(\operatorname N(\pi)\equiv1\pmod3\) by writing \(\pi=\frac{a+b\sqrt{-3}}2\).
Remark 8. Let’s run some checks on this definition. To see that \(\chi_\pi\) is unique, note that \(\pi\nmid3\) implies that the element of \(\mu_3\) are distinct in \(\mathbb Z[\omega]/(\pi)\): for example, \(1-\omega\) is nonzero because the norm of \(1-\omega\) is \(3\). On the other hand, the element \(\mu_3\subseteq(\mathbb Z[\omega]/(\pi))^\times\) are the unique three solutions to \(\beta^3=1\), so the well-definedness of \(\chi_\pi\) follows by noting \[\left(\alpha^{(\operatorname N(\pi)-1)/3}\right)^3\equiv1\pmod\pi.\] Lastly, we note that \(\chi_\pi\) is multiplicative because \((\cdot)^{(\operatorname N(\pi)-1)/3}\) is multiplicative.
Remark 9. By construction, \(\chi_\pi\) is trivial on the cubes of \((\mathbb Z[\omega]/(\pi))^\times\). On the other hand, let’s check that \(\chi_\pi\) is a nontrivial character. Indeed, because \(\pi\) is prime, we see that \((\mathbb Z[\omega]/(\pi))^\times\) is cyclic, so we may let \(g\) be a generator. Then \[g^{(\operatorname N(\pi)-1)/3}\not\equiv1\pmod{\pi}\] by definition of \(g\), so \(\chi_\pi(g)\ne1\) is forced.
The discussion preceding Definition 7 yields the following lemma.
Definition 10. Fix a prime \(p\). Given a character \(\chi\colon(\mathbb Z/p\mathbb Z)^\times\to\mathbb C^\times\), we define the Gauss sum \[g(\chi)=\sum_{i=1}^{p-1}\chi(i)\zeta^i.\]
Lemma 11. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[\sum_{i=1}^{p-1}\zeta_p^{i^3}=g(\chi_\pi)+\overline{g(\chi_\pi)}-1.\] Here, \(\chi_\pi\) is a character on \((\mathbb Z/p\mathbb Z)^\times\) via the inclusion \(\mathbb Z\hookrightarrow\mathbb Z[\omega]\).
Proof. Quickly, note that \(\chi_\pi\) yields a well-defined character on \((\mathbb Z/p\mathbb)^\times\) because the kernel of the ring homomorphism \[\mathbb Z\hookrightarrow\mathbb Z[\omega]\twoheadrightarrow\frac{\mathbb Z[\omega]}{(\pi)}\] is generated by \(\operatorname N_{\mathbb Q(\omega)/\mathbb Q}(\pi)=p\). Observe that this induces an embedding \(\mathbb Z/p\mathbb Z\to\mathbb Z[\omega]/(p)\), which must be an isomorphism because these two rings have the same size.
We now proceed with the proof of the lemma. Let \(H\subseteq(\mathbb Z/p\mathbb Z)^\times\) be the subgroup of cubes. Then on one hand, \[\sum_{i=1}^{p-1}\zeta_p^{i^3}=3\sum_{i\in H}\zeta_p^i,\] as at the end of the proof of Corollary 6. On the other hand, for \(i\in(\mathbb Z/p\mathbb Z)^\times\), we claim that \[1+\chi_\pi(i)+\overline{\chi_\pi(i)}\stackrel?=\begin{cases}3 & \text{if }i\in H, \\0 & \text{if }i\notin H.\end{cases}\] Let’s explain why this claim completes the proof: summing over \(i\) shows that \[\sum_{i=1}^{p-1}\zeta_p^i+g(\chi_\pi)+\overline{g(\chi_\pi)}=3\sum_{i\in H}\zeta_p^i,\] completing the proof of the lemma.
It remains to prove the claim. Surely if \(i\in H\), then \(i\) is a cube, so \(\chi_\pi(i)=1\) follows by definition. For \(i\notin H\), it is enough to show that \(\chi_\pi(i)\ne1\) because \(1+\omega+\overline\omega=0\). But \(\chi_\pi\) is nontrivial on \((\mathbb Z[\omega]/(\pi))^\times\) by Remark 9, so it is nontrivial on \((\mathbb Z/p\mathbb Z)^\times\). We already know that it is trivial on \(H\), so it descends to a nontrivial character on the three-element group \((\mathbb Z/p\mathbb Z)^\times/H\). However, all nontrivial characters on the three-element group are nontrivial on nontrivial elements, so the result follows. \(\blacksquare\)
Lemma 11 tells us that it will be convenient to instead compute the irreducible cubic for \[\sum_{i=0}^{p-1}\zeta_p^{i^3}=g(\chi_\pi)+\overline{g(\chi_\pi)}.\]
Lemma 12. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then the element \[G=\sum_{i=0}^{p-1}\zeta_p^{i^3}\] satisfies the equation \(G^3-3pG-2\operatorname{Re}g(\chi_\pi)^3=0\).
Proof. Set \(g=g(\chi_\pi)\) for brevity. Using Lemma 11, we simply compute \[\begin{aligned}G^3 &= (g+\overline g)^3 \\&= g^3+\overline g^3+3g\overline g(g+\overline g) \\&= 3g\overline gG+2\operatorname{Re}g^3.\end{aligned}\] The result follows after rearrangement as soon as we note that \(g\overline g=p\) as worked out in this post. \(\blacksquare\)
It remains to compute \(g(\chi_\pi)^3\). As evidenced by the statement of Theorem 1, this must be slightly complicated. Let’s start with some partial information.
Lemma 13. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[g(\chi_\pi)^3\equiv1\pmod3.\]
Proof. We are trying to show that \(g(\chi_\pi)^3-1\) is an algebraic integer divisible by \(3\). Using the Frobenius automorphism, we see that \[g(\chi_\pi)^3\equiv\sum_{i=1}^{p-1}\chi_\pi(i)^3\zeta_p^{3i}\pmod3.\] However, \(\chi_\pi(i)^3=1\) for all \(i\), so because \(3\) and \(p\) are coprime, we see that \[g(\chi_\pi)^3\equiv\sum_{i=1}^{p-1}\zeta_p^i\pmod3,\] and now the right-hand side is \(-1\). \(\blacksquare\)
It looks like we can use unique prime factorization in \(\mathbb Z[\omega]\) to pin down \(g(\chi_\pi)\) because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\) and \(g(\chi_\pi)^3\equiv1\pmod3\). However, this does not make sense because \(g(\chi_\pi)\) is not an element of \(\mathbb Z[\omega]\). Luckily, it is not hard to salvage this idea.
Lemma 14. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[g(\chi_\pi)^3\in p\mathbb Z[\omega].\]
Proof. This is fairly tricky. Because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\), it is enough to show that \(g(\chi_\pi)^2\in\overline{g(\chi_\pi)}\mathbb Z[\omega]\), which we will do by a direct calculation. Indeed, \[\begin{aligned}g(\chi_\pi)^2 &= \sum_{i,j=1}^{p-1}\chi_\pi(i)\chi_\pi(j)\zeta_p^{i+j} \\&= \sum_{k=0}^{p-1}\Bigg(\sum_{i+j=k}\chi_\pi(i)\chi_\pi(j)\Bigg)\zeta_p^k \\&= \sum_{i=1}^{p-1}\chi_\pi(i)\chi_\pi(-j)+\sum_{k=1}^{p-1}\Bigg(\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j)\Bigg)\chi_\pi(k)^2\zeta_p^k. \\&= \chi_\pi(-1)\sum_{i=1}^{p-1}\chi_\pi(i)^2+\Bigg(\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j)\Bigg)g\left(\chi_\pi^2\right).\end{aligned}\] Note that \(\chi_\pi^2=\overline{\chi_\pi}\) because \(\chi_\pi\) has codomain \(\mu_3\), so \(g\left(\chi_\pi^2\right)=\chi_\pi(-1)\overline{g(\chi_\pi)}\) follows after a rearrangement. Because \(\chi_\pi\) is nontrivial, it follows that the left sum vanishes, so the result follows because \(\operatorname{im}\chi_\pi\subseteq\mathbb Z[\omega]\). \(\blacksquare\)
Remark 15. In general, given nontrivial Dirichlet characters \(\chi_1,\chi_2\pmod p\) for which \(\chi_1\chi_2\) is nontrivial, the same argument shows that \[\frac{g(\chi_1)g(\chi_2)}{g(\chi_1\chi_2)}=\sum_{i+j=1}\chi_1(i)\chi_2(j).\] The right-hand sum is important enough to deserve its own name: it is called a Jacobi sum.
Proposition 16. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\) with \(\pi\equiv1\pmod3\). Then \[g(\chi_\pi)^3\in\{-p\pi,-p\overline\pi\}.\]
Proof. By Lemma 14, \(g(\chi_\pi)^3\in p\mathbb Z[\omega]\). Because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\), we see that the prime factorization \(g(\chi_\pi)^3/p\) is associate to either \(\pi\) or \(\overline\pi\). To finish, note that we can remove the associate ambiguity by Lemma 2 because \(g(\chi_\pi)^3/p\equiv-1\pmod3\) by Lemma 13. \(\blacksquare\)
We are now ready to collect everything we have done in this blog post.
Proof of Theorem 1. The first part of the theorem follows from Proposition 3 and Remark 4. In fact, Remark 4 has shown that \[\pi=\frac{A+3B\sqrt{-3}}2\] is the unique factor of \(p\) in \(\mathbb Z[\omega]\) with \(\operatorname{Im}\pi>0\) and \(\pi\equiv1\pmod3\).
We now turn to the second part of the theorem. By Corollary 6 and Lemma 12, we see that the cubic subfield is generated by an element \(G\) which is the root of the polynomial \[x^3-3px-2\operatorname{Re}g(\chi_\pi)^3.\] However, \(g(\chi_\pi)^3\in\{-p\pi,-p\overline\pi\}\) by Proposition 16, so \(-2\operatorname{Re}g(\chi_\pi)^3=pA\). \(\blacksquare\)
Remark 17. The element \(\sum_{i=1}^{p-1}\zeta_p^{i^3}\) will not generate the ring of integers in the cubic subfield because it is divisible by \(3\). However, \[\frac13\sum_{i=1}^{p-1}\zeta_p^{i^3}\] also may not generate the ring of integers. For example, if \(p=31\), then the discriminant of this element is divisible by \(4\); however, the discriminant of the cubic subfield must divide the discriminant of \(\mathbb Q(\zeta_p)\), and the latter is a power of \(p\). On the other hand, discriminant calculations are enough to verify that this element generates the ring of integers when \(p\in\{7,13,19,37\}\): in these cases, the discriminant of this element is \(p^2\), but \(p\) does ramify in the cubic subfield.
]]>Remark 18. It is possible to determine which of \(p\pi\) or \(p\overline\pi\) equals \(g(\chi_\pi)^3\) by computing the Jacobi sum \[\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j).\] The answer is slightly complicated; it is worked out as Theorem 3.1.3 in the book Gauss Sums and Jacobi Sums by Berndt, Evans, and Williams.
Proposition 1. Let \(f\) be a polynomial over a field \(k\). Then there is a polynomial \(g\) divisible by \(f\) which is a linear combination of the polynomials \[\{x^p:p\text{ is prime}\}.\]
Proof. Observe that \[\frac{k[x]}{(f(x))}\] is a finite-dimensional vector space. Thus, some nonzero linear combination of the polynomials in \[\{x^p:p\text{ is prime}\}.\] is required to vanish in the quotient \(k[x]/(f(x))\). The result follows. \(\blacksquare\)
]]>Remark 2. Of course, there is noting special about primes, as the proof makes clear.
We are after the following theorem.
Notation 1. Fix an extension \(L/K\) of number fields. Then we let \(\operatorname{Spl}(L/K)\) denote the collection of prime ideals of \(\mathcal O_K\) which are totally split in \(L\).
Theorem 2. Fix a Galois extension \(L/K\) of number fields, and assume that \(K\) is Galois over \(\mathbb Q\). If the set \(\operatorname{Spl}(L/K)\) is invariant under \(\operatorname{Gal}(K/\mathbb Q)\) (possibly away from a finite set), then \(L\) is Galois over \(\mathbb Q\).
Corollary 3. Fix a Galois extension \(K\) of \(\mathbb Q\). Then the Hilbert class field \(H\) of \(K\) is Galois over \(\mathbb Q\).
Proof. A prime \(\mathfrak p\) of \(K\) splits completely in \(H\) if and only if it is principal. However, being principal in \(K\) is a property invariant under \(\operatorname{Gal}(K/\mathbb Q)\), so the corollary follows from the theorem. \(\blacksquare\)
We now turn to the proof of Theorem 2. Our main input will be the following recognition result.
Proposition 4. Fix Galois extensions \(L\) and \(L’\) of a number field \(K\). Then \(L\subseteq L’\) if and only if \(\operatorname{Spl}(L’/K)\subseteq\operatorname{Spl}(L/K)\) (possibly away from a finite set).
The inclusion in the proposition takes place in a fixed algebraic closure of \(\mathbb Q\).
Proof. We show the directions separately.
Remark 5. We do not need the full power of the Chebotarev density theorem to show that the density of split primes in an extension \(L/K\) is \(1/[L:K]\). In short, the set of primes \(\mathfrak q\) in \(L\) which lie over a totally split prime of \(K\) has Dirichlet density \(1\) for norm reasons: not splitting completely increases inertial degree, which causes the norm to explode (namely, note \(\sum_p1/p^2<\infty\)). But then the Dirichlet density of totally split primes in \(K\) is \[\lim_{s\to1^+}\frac{\sum_{\mathfrak p\in\operatorname{Spl}(L/K)}\operatorname N(\mathfrak p)^{-s}}{\sum_{\mathfrak p}\operatorname N(\mathfrak p)^{-s}}=\lim_{s\to1^+}\frac{\frac1{[L:K]}(-\log(s-1))}{(-\log(s-1))},\] which is \(\frac1{[L:K]}\).
Observe that the proof of Proposition 4 more or less upgrades the following special case.
Corollary 6. Fix Galois extensions \(L\) and \(L’\) of a number field \(K\). Then \(L=L’\) if and only if \(\operatorname{Spl}(L’/K)=\operatorname{Spl}(L/K)\) (possibly away from a finite set).
Proof. This is the equality case of Proposition 4. \(\blacksquare\)
Proof of Theorem 2. Embed all fields into \(\overline{\mathbb Q}\). We would like to show that \(\sigma(L)=L\) for any automorphism \(\sigma\) of \(\overline{\mathbb Q}\). By Corollary 6, it is enough to check that \[\operatorname{Spl}(L/K)\stackrel?=\operatorname{Spl}(\sigma(L)/K),\] possibly away from a finite set. (Note \(\sigma(K)=K\) because \(K\) is Galois over \(\mathbb Q\).) Well, a prime \(\mathfrak p\) of \(K\) splits completely in \(\sigma(L)\) if and only if \(\mathrm{Frob}_{\mathfrak p}|_{\sigma(L)}\) is trivial. However, hitting everything in sight with \(\sigma^{-1}\), it is equivalent to show that \[\sigma^{-1}\mathrm{Frob}_{\mathfrak p}\sigma|_L\] is trivial, which is equivalent to \(\mathrm{Frob}_{\sigma^{-1}\mathfrak p}|_L\) being trivial. This last condition is equivalent to \(\sigma^{-1}\mathfrak p\in\operatorname{Spl}(L/K)\), which is equivalent to \(\mathfrak p\in\operatorname{Spl}(L/K)\) (possibly away from a finite set) by the Galois-invariance hypothesis. This completes the proof. \(\blacksquare\)
Now that we are done proving our theorem, let’s give another application.
Example 7. Let \(K\) be an imaginary quadratic field, and choose \(\alpha\) so that \(\mathcal O_K=\mathbb Z+\mathbb Z\alpha\). For each positive integer \(n\), we define the order \(\mathcal O_n=\mathbb Z+\mathbb Zn\alpha\), and we define the ring class field \(H_n\) to correspond to the adelic quotient \[K^\times\backslash\mathbb A_{K,f}^\times/\widehat{\mathcal O}^\times_n\] via class field theory. As such, away from a finite set, a prime \(\mathfrak p\) of \(K\) splits completely in \(H_n\) if and only if a uniformizer of \(K_{\mathfrak p}^\times\hookrightarrow\mathbb A_{K,f}^\times\) lands in \(K^\times\widehat{\mathcal O}^\times_n\). This latter condition is conjugation-invariant (because \(K^\times\) and \(\widehat{\mathcal O}^\times_n\) are both preserved by conjugation), so Theorem 2 implies that the extension \(H_n/\mathbb Q\) is Galois.
]]>Remark 8. Of course, Theorem 2 is not the only way to prove the claim of Example 7, but it is a quick way. Note that the “usual” argument which shows that the Hilbert class field is Galois does not really apply: one usually shows that the Hilbert class field is Galois by using the fact that it is the maximal unramified abelian extension, which is a property preserved by composites and embeddings. It seems trickier to give an analogous description of the ring class fields in Example 7.