Genus Theory

We compute the \(2\)-part of the class group of an imaginary quadratic field \(K\).

Theorem 1. Let \(K\) be an imaginary quadratic field, and let \(S\) be the set of primes of \(K\) which are ramified.

• Then \(\dim_{\mathbb F_2}\operatorname{Pic}(\mathcal O_K)[2]=\left|S\right|-1\).
• The vector space \(\operatorname{Pic}(\mathcal O_K)[2]\) is spanned by \(S\).

For example, it follows that \(K\) having unique prime factorization implies that \(K\in\{\mathbb Q(i),\mathbb Q(\sqrt{-2})\}\) or \(K=\mathbb Q(\sqrt{-p})\) where \(p\) is a prime with \(p\equiv3\pmod4\).

Remark 2. Of course, it is also possible to show the previous sentence more directly. Say \(K=\mathbb Q(\sqrt d)\) with \(d<0\), and let the discriminnt be \(D\); we want to show that \(D\) has only one prime factor. A principal ramified prime \(\mathfrak p\) over \((p)\) being principal provides an element with norm \(p\), which allows us to solve some quadratic form \(Q_K\) equal to \(p\). Expanding out \(Q_K\) as either \(x^2-dy^2\) or \(x^2+xy+\frac{1-d}4y^2\) shows that \[p\ge\frac14\left|D\right|.\] Thus, if \(D\) has more than one prime factor, then \(\left|D\right|\le16\), and one can handle these cases by hand.

Our exposition follows Matthew Emerton. Many of the arguments we will make will work for more general Galois extensions \(L/K\) of number fields. In this situation, genus theory is about understanding \(\operatorname{Pic}(\mathcal O_L)^{\operatorname{Gal}(L/K)}\), where \(G\) is the Galois group. To explain the relation to the \(2\)-part of the class group, we note the following.

Lemma 3. Let \(L/K\) be a quadratic extension of number fields with Galois group \(G\), and assume that \(\operatorname{Pic}\mathcal O_K\) is trivial. Then \[\operatorname{Pic}(\mathcal O_L)^{G}=\operatorname{Pic}(\mathcal O_L)[2].\]

Proof. Let \(\sigma\in G\) be the nontrivial element. Then we claim that \(\sigma[I]=-[I]\) for any ideal class \([I]\). This implies that \([I]\) is fixed by \(\sigma\) if and only if \([I]\) is \(2\)-torsion, thereby proving the lemma.

It remains to show the claim. By unique prime factorization, it is enough to show the claim in the case that \(I\) is a prime ideal \(\mathfrak q\), and we set \(\mathfrak p=\mathfrak q\cap K\) be the prime lying under \(\mathfrak q\). We now combine two observations.

• On one hand, because \(K\) has trivial class group, we see that \(\mathfrak p\) is principal.
• On the other hand, we claim that there is an integer \(r\) for which \[\mathfrak p^r\mathcal O_L\stackrel?=\mathfrak q\cdot\sigma\mathfrak q.\] This can be checked depending on the splitting behavior of \(\mathfrak p\) and the fact that \(\operatorname{Gal}(L/K)\) acts transitively on the primes lying above \(\mathfrak p\). Explicitly, if \(\mathfrak p\) is split or ramified, then \(r=1\); if \(\mathfrak p\) is inert, then \(r=2\).

Combining the above observations shows that \(\sigma\mathfrak q\) is the inverse of \(\mathfrak q\) in the class group. \(\blacksquare\)

We are now interested in studying \(\operatorname{Pic}(\mathcal O_L)^{\operatorname{Gal}(L/K)}\). Unsurprisingly, we will use Galois cohomology. The main point is to compute with the long exact sequence of \[0\to\operatorname{PDiv}(\mathcal O_L)\to\operatorname{Div}(\mathcal O_L)\to\operatorname{Pic}(\mathcal O_L)\to0.\] Thus, we will be interested in the cohomology of \(\operatorname{PDiv}(\mathcal O_L)\) and \(\operatorname{Div}(\mathcal O_L)\).

Lemma 4. Let \(L/K\) be a Galois extension of number fields with Galois group \(G\).

• There is a right exact sequence \[\operatorname{PDiv}(\mathcal O_K)\to\operatorname{PDiv}(\mathcal O_L)^G\to\mathrm H^1(G;\mathcal O_L^\times)\to0.\]
• There is a left exact sequence \[0\to\mathrm H^1(G;\operatorname{PDiv}(\mathcal O_L))\to\mathrm H^2(G;\mathcal O_L^\times)\to\mathrm H^2(G;L^\times).\]

Proof. Both claims follow from taking Galois cohomology of the short exact sequence \[1\to\mathcal O_L^\times\to L^\times\to\operatorname{PDiv}(\mathcal O_L)\to0.\] The right \(0\) and the left \(0\) in both claims arises by Hilbert’s theorem 90, which tells us \(\mathrm H^1(G;L^\times)\). Lastly, in the first claim, \(\operatorname{PDiv}(\mathcal O_K)\) is simply being used as a stand-in for the image of \(K^\times\) in \(\operatorname{PDiv}(\mathcal O_L)^G\). \(\blacksquare\)

Lemma 5. Let \(L/K\) be a Galois extension of number fields with Galois group \(G\).

• The group \(\mathrm H^0(G;\operatorname{Div}(\mathcal O_L))\) is free abelian generated by \(\mathfrak p^{1/e_{\mathfrak p}}\)s. Here, \(\mathfrak p\) is a prime ideal of \(\mathcal O_K\), and \(\mathfrak p^{1/e_{\mathfrak p}}\) is the unique ideal in \(\mathcal O_L\) whose \(e_{\mathfrak p}\)th power is \(\mathfrak p\).
• The group \(\mathrm H^1(G;\operatorname{Div}(\mathcal O_L))\) is trivial.

Proof. Both calculations will come from using unique prime factorization of ideals to simplify \(\operatorname{Div}(\mathcal O_L)\). In particular, as an abelian group, we know that \(\operatorname{Div}(\mathcal O_L)\) is generated by the prime ideals. To recover the Galois action, we note that \(G\) acts transitively on the collection of primes \(S_{\mathfrak p}\) of primes above any given prime \(\mathfrak p\) of \(K\). Thus, \[\operatorname{Div}(\mathcal O_L)=\bigoplus_{\mathfrak p}\mathbb Z[S_{\mathfrak p}].\] Letting \(D_{\mathfrak p}\) be the stabilizer of some prime above \(\mathfrak p\) (which is well-defined up to conjugation), we see that we have an isomorphism \[\operatorname{Div}(\mathcal O_L)\cong\bigoplus_{\mathfrak p}\mathbb Z[G/D_{\mathfrak p}].\] of \(\mathbb Z[G]\)-modules. We now handle each part separately.

• To compute the \(G\)-invariants, we see that the \(G\)-invariants of \(\mathbb Z[G/D_{\mathfrak p}]\) simply look like \(\sum_{gD_{\mathfrak p}\in G/D_{\mathfrak p}}a_{gD_{\mathfrak p}}(gD_{\mathfrak p})\) with all coefficients equal. Thus, this is isomorphic to \(\mathbb Z\) generated by \(\sum_{gD_{\mathfrak p}\in G/D_{\mathfrak p}}(gD_{\mathfrak p})\). Unwinding the isomorphism, we see that this generator corresponds to the ideal in \(\operatorname{Div}(\mathcal O_L)\) which is the product of all the primes above \(\mathfrak p\), which is the desired statement.
• To compute higher cohomology, note \(\mathbb Z[G/D_{\mathfrak p}]=\mathbb Z[G]\otimes_{\mathbb Z[D_{\mathfrak p}]}\mathbb Z\), so Shapiro’s lemma informs us that \[\mathrm H^i(G;\operatorname{Div}(\mathcal O_L))=\bigoplus_{\mathfrak p}\mathrm H^i(D_{\mathfrak p};\mathbb Z).\] For \(i=1\), the summands are \(\mathrm{Hom}(D_{\mathfrak p},\mathbb Z)\), which is trivial. \(\blacksquare\)

Proposition 6. Let \(L/K\) be a Galois extension of number fields with Galois group \(G\). Then there is an exact sequence \[0\to\mathrm H^1(G;\mathcal O_L^\times)\to\frac{\operatorname{Div}(\mathcal O_L)^G}{\operatorname{PDiv}(\mathcal O_K)}\to\operatorname{Pic}(\mathcal O_L)^G\to\mathrm H^2(G;\mathcal O_L^\times)\to\mathrm H^2(G;L^\times).\]

Proof. There is a short exact sequence \[0\to\operatorname{PDiv}(\mathcal O_L)\to\operatorname{Div}(\mathcal O_L)\to\operatorname{Pic}(\mathcal O_L)\to0.\] Taking group cohomology gives \[0\to\operatorname{PDiv}(\mathcal O_L)^G\to\operatorname{Div}(\mathcal O_L)^G\to\operatorname{Pic}(\mathcal O_L)^G\to\mathrm H^1(G;\operatorname{PDiv}(\mathcal O_L))\to0,\] where the last \(0\) follows from Lemma 4. We now fix this exact sequence.

• By Lemma 3, we may replace the ending surjection of this exact sequence with \[\operatorname{Pic}(\mathcal O_L)^G\to\mathrm H^2(G;\mathcal O_L^\times)\to\mathrm H^2(G;L^\times).\]
• Because \(\operatorname{PDiv}(\mathcal O_K)\) vanishes in \(\operatorname{Pic}(\mathcal O_L)\), we may replace the start of the exact sequence with \[0\to\frac{\operatorname{PDiv}(\mathcal O_L)^G}{\operatorname{PDiv}(\mathcal O_K)}\to\frac{\operatorname{Div}(\mathcal O_L)^G}{\operatorname{PDiv}(\mathcal O_K)}\to\operatorname{Pic}(\mathcal O_L)^G.\] The result now follows Lemma 3. \(\blacksquare\)

The moral is that we have managed to compute \(\operatorname{Pic}(\mathcal O_L)^G\) in terms of the cohomology of \(\mathcal O_L^\times\) and \(L^\times\). Let’s apply this to the case of an imaginary quadratic field, proving Theorem 1.

Lemma 7. Let \(K\) be an imaginary quadratic extension of \(\mathbb Q\) with Galois group \(G\).

• The group \(\mathrm H^1(G;\mathcal O_K^\times)\) has two elements.
• The map \(\mathrm H^2(G;\mathcal O_K^\times)\to\mathrm H^2(G;K^\times)\) is injective.

Proof. Because \(G\) is cyclic, we may let \(\sigma\in G\) be a generator. Recall that the cohomology of any \(G\)-module \(M\) is given by \[\mathrm H^i(G;M)=\begin{cases} \ker{\operatorname N_G}|_M/(\sigma-1)M & \text{if }i\text{ is odd}, \\{} M^G/\operatorname N_G(M) & \text{if }i\text{ is even}. \end{cases}\] for \(i>0\); further, these isomorphisms are functorial in \(M\).

• For the first claim, we note the kernel of the norm on \(\mathcal O_K^\times\) is all of \(\mathcal O_K^\times\): all units are roots of unity, and applying complex conjugation yields the inverse of a root of unity. It follows that \(\sigma\) acts by \(-1\) on \(\mathcal O_K^\times\), so the quotient \[\mathrm H^1(G;\mathcal O_K^\times)=\frac{\mathcal O_K^\times}{(\sigma-1)\mathcal O_K^\times}\] is just \(\mathcal O_K^\times/\mathcal O_K^{\times2}\). Because \(\mathcal O_K^\times\) is cyclic (after all, it is a group of roots of unity), it follows that the quotient has size \(2\).
• For the second claim, we would like to show that the map \[\frac{\mathbb Z^\times}{\operatorname N_{K/\mathbb Q}(\mathcal O_K^\times)}\to\frac{\mathbb Q^\times}{\operatorname{N}_{K/\mathbb Q}(K^\times)}\] is injective. Well, the left-hand side is just \(\{\pm1\}\), so this amounts to nothing that \(-1\) is not the image of a norm because norms from imaginary quadratic fields are positive. \(\blacksquare\)

Proof of Theorem 1. Let the Galois group of the extension \(K/\mathbb Q\) be \(G\). By plugging in the results of Lemmas 3 and 7 into the exact sequence of Proposition 6, we get a short exact sequence \[0\to\mathbb F_2\to\frac{\operatorname{Div}(\mathcal O_K)^G}{\operatorname{PDiv}(\mathbb Q)}\to\operatorname{Pic}(\mathcal O_K)[2]\to0.\] The middle quotient was computed in Lemma 5, which tells us that it is free abelien generated by \((p)^{1/e_p}\) for each rational \(p\) lying under a prime in \(S\). Of course, \((p)^{1/e_p}\) is the unique prime in \(S\) lying over \((p)\), so we see that the middle quotient is the free vector space over \(\mathbb F_2\) spanned by \(S\). Both claims now follow from the exact sequence. \(\blacksquare\)

Remark 7. There is an analogous statement as Theorem 1 for real quadratic fields, but it is slightly more complicated. (However, I suspect that Remark 2 does not generalize.) For example, the map \[\mathrm H^2(G;\mathcal O_K^\times)\to\mathrm H^2(G;K^\times)\] does not have to be injective, so \(\operatorname{Pic}(\mathcal O_K)[2]\) need not be generated by the ramified primes. Luckily, the discrepancy is not too significant because \(\mathrm H^2(G;\mathcal O_K^\times)\) has size at most \(2\).