Today I Learned

June 29th

Today I learned more about the Zariski topology. To review, here is the definition in the specialized case of interest.

We remark that, for any $\{f_\ell\}_{\ell=1}^m\subseteq S$ and $\{a_\ell\}_{\ell=1}^m\subseteq k,$ we have that $a\in V(S)$ implies\[\sum_{\ell=1}^ma_\ell f_\ell(a)=\sum_{\ell=1}^ma_\ell\cdot0=0\]for free. In other words, adding in $k$-linear combinations of elements of $S$ does not limit the vanishing set any more $V(S).$ Because adding in more functions certainly cannot make the vanishing set larger (the condition is "for all''), we conclude that $V(S)=V((S)),$ where $(S)$ is the ideal of $k[t_1,\ldots,t_n]$ generated by $S.$

Anyways, it is not difficult to verify that this actually makes a topology, which we do now.

To show that some family of closed sets makes a topology, we have to show the following.

• The empty set is an algebraic set. For example, the constant polynomial $1\in k[t_1,\ldots,t_n]$ doesn't vanish anywhere, so $V(\{1\})=\emp.$

• The entire set $\AA_k^n$ is an algebraic set. Indeed, the constant polynomial $0\in k[t_1,\ldots,t_n]$ vanishes everywhere, so $V(\{0\})=\AA_k^n.$

• Algebraic sets are closed under finite union. More precisely, we need to show that, for any two algebraic sets $V(S_1)$ and $V(S_2),$ then $V(S_1)\cup V(S_2)$ is also algebraic. We claim that \[V(S_1)\cup V(S_2)\stackrel?V(S_1S_2),\] where $S_1S_2:=\{f_1f_2:f_1\in S_1,f_2\in S_2\}.$ In one direction, note that $a\in V(S_1)\cup V(S_2)$ lives in $V(S_1)$ without loss of generality. Thus, for any $f_1f_2\in S_1S_2$ with $f_1\in S_1$ and $f_2\in S_2,$ we see \[(f_1f_2)(a)=f_1(a)f_2(a)=0f_2(a)=0\] because $a\in V(S_1)$ implies $f_1(a)=0.$

  In the other direction, fix $a\in V(S_1S_2).$ If $a\in V(S_1),$ then we are done already. Otherwise, $a\notin V(S_1),$ so there is some $f_1\in S_1$ such that $f_1(a)\ne0.$ We want to show $a\in V(S_2).$ Well, for any $f_2\in V(S_2),$ we know $f_1f_2\in S_1S_2,$ so it follows \[f_1(a)f_2(a)=(f_1f_2)(a)=0\] because $a\in V(S_1S_2).$ However, $f_1(a)\ne0,$ so $f_2(a)=0$ because we live in an integral domain (!). Thus, $a\in V(S_2).$

• Algebraic sets are closed under arbitrary intersection. For this, we claim that, given a collection $\{S_\alpha\}_{\alpha\in\lambda}$ with $S_\alpha\subseteq k[t_1,\ldots,t_n],$ we have \[\bigcap_{\alpha\in\lambda}V(S_\alpha)=V\left(\bigcup_{\alpha\in\lambda}S_\alpha\right).\] Showing this is a matter of unravelling the definitions. The set on the left expands as \[\bigcap_{\alpha\in\lambda}\left\{a\in\AA_k^n:f(a)=0\text{ for each }f\in S_\alpha\right\}.\] The set on the right expands as \[\left\{a\in\AA_k^n:f(a)=0\text{ for each }f\in\bigcup_{\alpha\in\lambda}S_\alpha\right\}.\] These set constructions are saying the same thing: we want $a\in\AA_k^n$ such that $a$ vanishes on any $f\in S_\alpha$ for any $S_\alpha.$ So these are equal.

Having verified all properties needed to make a topology on closed sets, we are done here. $\blacksquare$

Today we are interested in more topological properties of $\AA_k^n.$ However, we understand $\AA_k^n$ best as some algebraic object, so we will slowly try to turn algebraic definitions into statements about the topology. This means that we will want a way to turn our closed sets into algebraic objects, namely ideals. We define\[I(V):=\{f\in k[t_1,\ldots,t_n]:f(a)=0\text{ for each }a\in V\}\]for any algebraic set $V.$ Because algebraic sets $V$ are defined as $V(S)$ for some $S\subseteq k[t_1,\ldots,t_n],$ this definition is somewhat natural as an attempt to define an inverse.

Quickly, it is true that $I(V)$ is an ideal. Indeed, $0\in I(V),$ and if $a_1,a_2\in k[t_1,\ldots,t_n]$ and $f_1,f_2\in I(V)$ has $a_1f_1+a_2f_2\in I(V)$: for any $a\in V,$ we know\[(a_1f_1+a_2f_2)(a)=a_1f_1(a)+a_2f_2(a)=0.\]So $I(V)$ is closed under $k[t_1,\ldots,t_n]$-linear combination and is an ideal.

It turns out that $I(-)$ is not actually an inverse of $V(-).$ For example, if $S$ is not an ideal, we already remarked that $S$ and $(S)$ generated the same vanishing set, so it is impossible for $I(-)$ to distinguish between these. We can say a bit more than this.

Fix $f\in\sqrt{(S)}$ so that we want to show that $f\in I(V(S)).$ By definition,\[I(V(S))=\left\{g\in k[t_1,\ldots,t_n]:g(a)=0\text{ for each }a\in V(S)\right\},\]so we need to show that $f$ satisfies $f(a)=0$ for each $a\in V(S).$

Well, $f\in\sqrt{(S)}$ promises a positive integer $N$ such that $f^N\in(S),$ which promises finite sequences $\{a_\ell\}_{\ell=1}^m\subseteq k$ and $\{f_\ell\}_{\ell=1}^m$ such that\[f^N=\sum_{\ell=1}^ma_\ell f_\ell.\]Finishing up now, we can pick up any $a\in V(S),$ and we note that\[f^N(a)=\sum_{\ell=1}^ma_\ell f_\ell(a)=0\]because the $f_\bullet\in S.$ However, $f(a)^N=0$ forces $f(a)=0$ because we live in an integral domain. Thus, we do indeed have $f\in I(V(S)),$ which finishes the proof. $\blacksquare$

In fact, the above proposition can be strengthened into an equality in most cases we care about, but this requires quite a bit more effort; it is Hilbert's Nullstellensatz. We will not dwell on this further because we have geometry to get back to.

Quickly, we pick up the following.

This is a matter of expanding out our definitions. To start, if $S_1\subseteq S_2\subseteq k[t_1,\ldots,t_n],$ pick up $a\in V(S_2)$ so that we need to show $V(S_1).$ Well, for any $f\in S_1,$ we know $f\in S_2,$ so it follows\[f(a)=0.\]It follows $a\in V(S_1),$ so $V(S_2)\subseteq V(S_1)$ follows. We note that we don't have the converse here because $S_1$ and $S_2$ are not necessarily behaved as sets, or even as ideals; e.g., $V((t))=V\left(\left(t^2\right)\right)$.

Similarly, fix algebraic sets $V_1\subseteq V_2$ so that we want $I(V_2)\subseteq I(V_1).$ Well, for any $f\in I(V_2),$ we know that any $a\in V_1$ also has $a\in V_2,$ so\[f(a)=0\]because $f\in I(V_2).$ Thus, $f\in I(V_1),$ so $I(V_2)\subseteq I(V_1)$ indeed.

As for the converse, suppose $I(V_2)\subseteq I(V_1)$ so that we want $V_1\subseteq V_2.$ Well, we may write $V_2=V(S_2)$ for some $S_2\subseteq k[t_1,\ldots,t_n],$ and because $V_2$ vanishes on $S_2,$ we see\[S_2\subseteq I(V_2)\subseteq I(V_1).\]Thus, any $a_1\in V_1,$ which vanishes on all of $I(V_1),$ will vanish on $S_2,$ which gives the needed $a_1\in V_2.$ This finishes the proof.$\blacksquare$

Now, primality is a condition we care about, so it would be nice to have a geometric condition on the closed set $V$ which gives $I(V)$ prime. Well, a proper ideal $\mf p$ is prime if and only if $ab\in\mf p$ implies $a\in\mf p$ or $b\in\mf p.$ A more ideal-theoretic way to say this is that $IJ\subseteq\mf p$ implies $I\subseteq\mf p$ or $J\subseteq\mf p.$

Translating this into vanishing sets means we want to make all ideals into closed sets and reverse the inclusions. This gives the following definition.

We might be tempted to say $X\subseteq X_1$ or $X\subseteq X_2$ is a more natural conclusion, but these closed sets would be induced by the subspace topology, so we get $X_1,X_2\subseteq X$ for free.

Intuitively, irreducible is essentially a very strong version of connected; after all, the definition of irreducible is the definition of connected where $X_1\sqcup X_2$ is replaced with $X_1\cup X_2.$ Explicitly, we have the following.

Well, to verify connectivity, suppose $X=X_1\sqcup X_2$ for disjoint closed sets $X_1$ or $X_2.$ Then $X=X_1\cup X_2,$ so $X_1=X$ or $X_2=X.$ Without loss of generality, $X_1=X,$ so $X_2\subseteq X\setminus X_1=\emp$ because $X_1$ and $X_2$ are disjoint, giving $X_2=\emp.$ It follows $X$ is connected. $\blacksquare$

However, it turns out that being irreducible is a rather strong condition.

In fact, we have the following.

We show that a Hausdorff space $X$ with at least two points is not irreducible. We have to use the Hausdorff condition somehow, so pick up our two distinct points $x_1,x_2\in X.$ Then because $X$ is Hausdorff, we are granted two disjoint open sets $U_1$ and $U_2$ such that $x_1\in U_1$ and $x_2\in U_2.$

Now, set\[X_1=X\setminus U_1\qquad\text{and}\qquad X_2=X\setminus U_2.\]We know that $X_1$ and $X_2$ are closed because they are the complement of open sets, and $U_1\cap U_2=\emp$ implies\[X=X\setminus\emp=X\setminus(U_1\cap U_2)=(X\setminus U_1)\cup(X\setminus U_2),\]so $X=X_1\cup X_2.$ However, we know $X\ne X_1$ and $X\ne X_2$ because $X\setminus X_1=U_1$ and $X\setminus X_2=U_2$ are both nonempty. It follows that $X$ is not irreducible. $\blacksquare$

It follows that, if irreducibility is a condition we care about, most of the spaces we are looking at aren't going to be Hausdorff. Granted, our topology wasn't made to be pretty; it was made to work.

Anyways, we introduced irreducible spaces to be the topological version of prime. Here is the codification of this.

The closed set $V$ is irreducible if and only if $V$ is irreducible under the subspace topology. Because our closed sets look like $V_\bullet\cap V$ for some other closed $V_\bullet,$ we can get rid of this intersection and say that our irreducible condition is asserting\[V\subseteq V_1\cup V_2\implies V\subseteq V_1\text{ or }V\subseteq V_2,\]where $V_1,V_2$ are algebraic sets themselves. Anyways, we need to show $I(V)$ is prime. Well, working element-wise, suppose $f_1f_2\in I(V)$ so that we need $f_1\in I(V)$ or $f_2\in I(V).$ To use irreducibility, we fix the algebraic sets\[V_1:=V(\{f_1\})\qquad\text{and}\qquad V_2=V(\{f_2\}).\]In our proof that vanishing sets form a topology, we the union of vanishing sets is the vanishing set on the product, so $V_1\cup V_2=V(\{f_1f_2\}).$

Now, we want $V\subseteq V_1\cup V_2$ to use irreducibility. Indeed, any $a\in V$ vanishes on $f_1f_2$ because $f_1f_2\in I(V),$ implying $a\in V(\{f_1f_2\}=V_1\cup V_2.$ Thus, irreducibility of $V$ now kicks in to tell us $V\subseteq V_1$ or $V\subseteq V_2.$ Without loss of generality, we take $V\subseteq V_1.$

We hope that this corresponds to $f_1\in I(V),$ which will finish this direction. Indeed,\[V\subseteq V_1=V(\{f_1\})\]implies that, for each $a\in V,$ we know $a\in V_1,$ so $f_1(a)=0.$ Thus, all of $V$ vanishes under $f_1,$ so $f_1\in I(V).$ This completes the proof of this direction.

In the other direction, suppose that $I(V)$ is a prime ideal, and we need to show that $V$ is irreducible. Well, suppose we can decompose $V=V_1\cup V_2$ where $V_1$ and $V_2$ are closed subsets of $V.$ Expanding our view as before, we are really supposing $V\subseteq V_1\cup V_2$ for $V_1$ and $V_2$ algebraic subsets of $\AA_k^n,$ and we need $V\subseteq V_1$ or $V\subseteq V_2.$

Taking $I(-),$ we know\[I(V)\supseteq I(V_1\cup V_2).\]We want to turn the right-hand side into a product of ideals to use the primality condition, so we pick up the following lemma.

It suffices to show that any generator $f_1f_2$ of $I(V_1)I(V_2)$ where $f_1\in I(V_1)$ and $f_2\in I(V_2)$ lives in $I(V_1\cup V_2).$ Well, for any $a\in V_1\cup V_2,$ we know $a\in V_1$ or $a\in V_2,$ so\[f_1(a)=0\text{ or }f_2(a)=0.\]Thus, $(f_1f_2)(a)=0,$ so indeed, $f_1f_2\in I(V_1\cup V_2).$ This finishes. $\blacksquare$

So we see $I(V)\supseteq I(V_1)I(V_2).$ Looking ideal-theoretically at the primality of $I(V),$ it follows $I(V_1)\subseteq I(V)$ or $I(V_2)\subseteq I(V).$ Thus, we get $V\subseteq V_1$ or $V\subseteq V_2$ by inclusion-reversing again, which completes the proof that $V$ is irreducible. $\blacksquare$

To reiterate what just happened, we codify the intuition as follows.

This is so important that irreducible algebraic sets is called a variety, though we will not use this terminology.

We will now move on from primality. Noetherian is another nice algebraic condition, so we translate it into geometry. Well, Noetherian asserts that any ascending chain of ideals\[I_1\subseteq I_2\subseteq I_3\subseteq\cdots\]eventually stabilizes with $I_\ell=I_N$ for each $\ell\ge N$ for some positive integer $N.$ When we want to talk about geometric, we turn ideals into closed sets and reverse the inclusions. So we have the following definition.

If we wanted to define Artinian in a similar way, I think we could, but we won't have as much use for those.

Noetherian is a pretty strong condition on topological spaces. For example, we have the following.

Luckily, the space we are most interested in is typically Noetherian.

Without much else to do, fix a descending chain of closed (i.e., algebraic) sets\[V_1\supseteq V_2\supseteq V_3\supseteq\cdots.\]We want to turn this into a chain of ideals in order to use the Noetherian condition. Because each of the $V_\bullet$ is algebraic, we are promised subsets $S_\bullet\subseteq k[t_1,\ldots,t_n]$ such that $V_\bullet=V(S_\bullet).$ Because $V(S_\bullet)=V((S_\bullet)),$ we may replace each $S_\bullet$ with the ideal generated by $S_\bullet.$

So we have a sequence of ideals $\{S_\ell\}_{\ell=1}^\infty$ generated from the $V_\bullet.$ However, this sequence is not necessarily ascending because, for example, we can have $V((t))\subseteq V\left(\left(t^2\right)\right),$ but $(t)$ is not a subset of $\left(t^2\right).$ Instead, we define\[I_\ell=\sum_{j=1}^\ell S_\ell\]to ensure we are ascending. Indeed, this definition gives $I_{\ell+1}=I_\ell+S_{\ell+1},$ so $I_\ell\subseteq I_{\ell+1}.$

This might make us nervous, but we do still have $V_\bullet=V(I_\bullet).$ Indeed, $S_\ell\subseteq I_\ell$ gives $V_\ell=V_\bullet(S_\ell)\supseteq V(I_\ell)$ already. As for the other direction, pick up $a\in V_\ell.$ Then $a\in V_j\supseteq V_\ell$ for each $j\le\ell$ because our chain is descending so\[a\in V(S_j)=V_j.\]The point is that, for any $f_j\in S_j$ for any $j\le\ell,$ we still get $f_j(a)=0.$ Converting this, for any $\{f_j\}_{j=1}^\ell$ with $f_j\in S_j,$ we see\[\sum_{j=1}^\ell f_j(a)=\sum_{j=1}^\ell0=0.\]Thus, $a\in V(S_1+S_2+\cdots+S_\ell)=V(I_\ell),$ completing the proof that $V_\bullet=V(I_\bullet).$

We are now ready to finish. Finally using the Noetherian condition on $k[t_1,\ldots,t_n],$ we see that the chain of ideals\[I_1\subseteq I_2\subseteq I_3\subseteq\cdots\]must stabilize. That is, there is a positive integer $N$ for which $I_k=I_N$ for each $k\ge N.$ But then $V_k=V(I_k)=V(I_N)=V_N$ for each $k\ge N$ as well, so we see that the chain of closed sets also stabilizes. This finishes the proof. $\blacksquare$

In particular, when $k$ is a field, we get that $k[t_1,\ldots,t_n]$ is Noetherian by Hilbert's basis theorem. So in this case $\AA_k^n$ is indeed a Noetherian space. I think I can imagine a converse, where Noetherian spaces induce Noetherian rings, but we would need to have a process to turn (say, Zariski) spaces back into rings, which we don't have yet.

Anyways, the proof of the proposition dealt in more generality than $k$ fields in order to make clear how the Noetherian condition on our ring gave the Noetherian space. So here is the idea.

There is of course other stuff we can do, but I think that's enough for today.