# Exercise 5.5.A. of Vakil. Support of $f$ is either empty, the origin, or all space.

The following is a question from Professor Vakil's notes on Algebraic Geometry.
$$\DeclareMathOperator{\Spec}{Spec}$$ $$\DeclareMathOperator{\Supp}{Supp}$$

Suppose $$f$$ is a function on $$\Spec \mathsf{k}[x, y]/(y^2, xy)$$ (i.e., $$f \in \mathsf{k}[x, y]/(y^2, xy))$$. Show that $$\Supp f$$ is either the empty set, or the origin, or the entire space.

This has been asked before here and here.
I did not completely understand the answers there. (Why do they look at $$A_x$$?)
I have a proof of my own that I would like to get verified.

Let $$A = \mathsf{k}[x, y]/(y^2, xy)$$.

Recall that $$\Supp f$$ is a closed subset of $$\Spec A$$. Also recall that the primes of $$\Spec A$$ are of the form $$[(y)]$$ or $$[(x - a, y)]$$ for some $$a \in \mathsf{k}.$$ Since all ideals of the latter form contain $$(y),$$ they are in its closure. Thus, if $$[(y)] \in \Supp f,$$ then $$\Supp f = \Spec A.$$

The above tells us that to prove the statement, it suffices to prove the following:
Suppose $$[(x - a, y)] \in \Supp f$$ for some $$a \in \mathsf{k}^\times.$$ Then, $$[(y)] \in \Supp f.$$

Now, in our case, the stalks are simply localisations at those primes and the map taking a function to its germ is simply the natural localisation map. Support that we are given $$f(x, y) \in A.$$ Then, since we have quotient-ed out $$(xy, y^2),$$ we can write $$f$$ as $$\begin{equation} \tag{*} \label{eq} f(x, y) = g(x) + cy \end{equation}$$ for some $$c \in \mathsf{k}.$$

Now, suppose that $$f(x, y)$$ is nonzero in the localisation at $$(x - a, y)$$ for $$a \neq 0.$$ (In other words, $$[(x - a, y)] \in \Supp f.$$)
Since $$x \notin (x - a, y),$$ we see that $$x$$ is a unit in the localisation. But $$xy = 0$$ in $$A$$ forces $$y = 0$$ in the localisation. Thus, \ref{eq} tells us that $$g(x) \neq 0.$$

We now claim that $$[(y)] \in \Supp f$$ as well.
As before, the $$cy$$ terms becomes $$0$$ in the localisation. A typical element of $$A \setminus (y)$$ looks like $$h(x) + by$$ for some $$b \in \mathsf{k}$$ and $$h(x) \neq 0.$$ Then, we have $$\begin{equation*} [h(x) + by]g(x) = h(x)g(x) + byg(0). \end{equation*}$$

Since $$h$$ and $$g$$ are nonzero, we see that the above is nonzero and thus, the image of $$f$$ is nonzero in the localisation and $$[(y)] \in \Supp f,$$ as desired.

(One can check that all the three possibilities are indeed attained. Consider $$0,$$ $$y,$$ and $$1$$ to get the support as $$\emptyset,$$ $$[(x, y)],$$ and $$\Spec A,$$ respectively.)

Two questions:

1. Is that correct?
2. Is there something easier that I'm missing?

This looks okay as long as you're assuming that $$k$$ is algebraically closed (I did not see this assumption in the problem): otherwise there are points of $$X$$ which are not of the form $$(x-a,y)$$. Consider $$k=\Bbb R$$ and $$(x^2+1,y)$$, for instance. While it is correct except for that hurdle, it feels overwrought to me and I think there is an easier way. The solution in your second link is in my opinion a great argument to use, so I'll attempt to add a bit of explanation for you.
Let's start with a preliminary fact: if $$a\in A$$ is a function on $$\operatorname{Spec} A$$ with $$A$$ an integral domain, then $$a$$ has support either $$\varnothing$$ or $$\operatorname{Spec} A$$. Proof: if $$a=0$$ in some local ring of $$\operatorname{Spec} A$$, then $$a=0$$ in $$A$$ because localization is injective for an integral domain.
Now we can see why they look at $$A_x\cong k[x]_x$$: taking our original function $$f$$ and restricting it to $$D(x)=\operatorname{Spec} k[x]_x\subset X$$, we see that our preliminary fact applies to $$f|_{D(x)}$$, and so $$\operatorname{Supp} (f|_{D(x)}) = D(x)\cap \operatorname{Supp} f$$ is either empty or all of $$D(x)$$. Therefore $$\operatorname{Supp} f$$ is a closed set either disjoint from or containing $$D(x)\subset X$$, which gives exactly the options listed.
• Oh yes, I certainly don't want to restrict myself to algebraically closed fields. I just want to confirm if that's how you are finally concluding: $A$ has only one minimal prime and thus $\operatorname{Spec} A$ is irreducible which means that $D(x)$ is dense. So if the support contains $D(x)$, it is the complete space. Is this the argument you had in mind? OTOH, if $D(x)$ is disjoint, then the support is contained in its complement $V(x) = \{(x, y)\}$. Jun 8, 2021 at 14:27
• @AryamanMaithani We know $\operatorname{Supp}(f|_{D(x)})$ is either $\varnothing$ or $D(x)$. We have $\operatorname{Supp} (f|_{D(x)}) = D(x)\cap \operatorname{Supp} f$. If $\operatorname{Supp} (f|_{D(x)}) = D(x)\cap \operatorname{Supp} f=\varnothing$, then $D(x)$ and $\operatorname{Supp} f$ are disjoint which means $\operatorname{Supp} f$ is either $\varnothing$ or $\{(x,y)\}$. If $\operatorname{Supp} (f|_{D(x)}) = D(x)$, then $D(x) \subset \operatorname{Supp} f$, hence $\operatorname{Supp} f$ is the whole space $\operatorname{Spec} k[x, y]/(y^2, xy)$. Jun 8, 2021 at 17:49
• @AryamanMaithani I was using irreducibility which is very similar to your density argument. Two ways I was thinking of are: i.) Since $D(x) \subset \operatorname{Supp} f$, then $\operatorname{Spec} k[x, y]/(y^2, xy) = \operatorname{Supp} f \cup \{(x,y)\}$, but since $\operatorname{Spec} k[x, y]/(y^2, xy)$ is irreducible, then $\operatorname{Supp} f=\operatorname{Spec} k[x, y]/(y^2, xy)$. Jun 8, 2021 at 20:20
• @AryamanMaithani ii.)$D(x) \subset \operatorname{Supp} f$ implies $\overline{D(x)} \subset \operatorname{Supp} f$, but since $D(x)$ is dense, then $\overline{D(x)} = \operatorname{Spec} k[x, y]/(y^2, xy)$; hence, $\operatorname{Supp} f = \operatorname{Spec} k[x, y]/(y^2, xy)$. Jun 8, 2021 at 20:20