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Let $f:X\to Y$ be a morphism of schemes, let $\mathcal{K}$ be the kernel of the structure map $\mathcal{O}_Y\to f_*\mathcal{O}_X$. Do we have

$$\mathrm{Supp}(\mathcal{O}_Y/\mathcal{K})=\overline{f(X)}?$$

Recall that the support of a sheaf (of abelian groups) is the set of points at which the stalk of that sheaf is nonzero.

I can prove $\subset$ as follows: If $y\in Y$ is in the support, then $(f_*\mathcal{O}_X)_y\neq 0$ (since the image of $1$ is $1$), so for any open neighborhood $V$ of $y$, we have $\mathcal{O}_X(f^{-1}V)\neq 0$, so $f^{-1}V\neq \emptyset$, thus $y\in \overline{f(X)}$. How about the convese? As $\mathcal{O}_Y/\mathcal{K}$ is of finite type, the support is closed, so one possible way is to show $f(X)\subset\mathrm{Supp}(\mathcal{O}_Y/\mathcal{K})$.

Somebody told me it's always true but you can add some mild conditions if you need.

Lao-tzu
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2 Answers2

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I think I can prove "=" in one stroke (without extra assumption), in detail as follows: the key observation is that any ring hom maps $0$ to $0$ and $1$ to $1$, and for the structure sheaf of a scheme $X\neq \emptyset$, we have $1\neq 0\in\mathcal{O}_X(U)\Leftrightarrow \mathcal{O}_X(U)\neq 0\Leftrightarrow U\neq \emptyset$. Thus $$y\in \mathrm{Supp}(\mathcal{O}_Y/\mathcal{K})\Leftrightarrow (\mathcal{O}_Y/\mathcal{K})_y\neq 0\Leftrightarrow (f_*\mathcal{O}_X)_y\neq 0\Leftrightarrow 1\neq 0\in(f_*\mathcal{O}_X)_y\Leftrightarrow 1\neq 0\in(f_*\mathcal{O}_X)(V), \forall V\Leftrightarrow \mathcal{O}_X(f^{-1}V)\neq 0, \forall V\Leftrightarrow f^{-1}V\neq \emptyset, \forall V\Leftrightarrow V\cap f(X)\neq \emptyset, \forall V\Leftrightarrow y\in \overline{f(X)},$$ where $\forall V$ means for all open neighborhoods $V$ of $y$ in $Y$; the 2nd $\Leftrightarrow$ is because we have an injective ring hom $(\mathcal{O}_Y/\mathcal{K})_y\rightarrow(f_*\mathcal{O}_X)_y$; the 4th $\Leftrightarrow$ is because the stalk is a filtred colimit over all open neighborhoods. Thus $$\mathrm{Supp}(\mathcal{O}_Y/\mathcal{K})=\overline{f(X)}.$$

Edit: According to the answer and comments below, the stated result is actually true for all morphisms of locally ringed spaces (but not for morphisms of ringed spaces), with the same argument.

Lao-tzu
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  • If you find an unreasonable point, please make a comment. – Lao-tzu Jul 30 '18 at 23:45
  • I have a question. Why the 2nd $\Leftarrow$ is true? – Plantation May 23 '22 at 10:39
  • For a injective ring homomorphism $f : A\to B$, although $B\neq 0 $, is there a possibility that $A =0$? – Plantation May 23 '22 at 10:43
  • @Plantation Since the structure map $\mathcal{O}Y\to f*\mathcal{O}_X$ is a morphism of sheaves of rings, $0$ goes to $0$ and $1$ goes to $1$. If $0\ne 1$ in the target, then same holds in the source. – Lao-tzu May 23 '22 at 16:58
  • Some confusion. You mean, $(\mathcal{O}Y)_y \to (f{}\mathcal{O}X)_y$ sends $0$ to $0$ and $1$ to $1$? And if $0 \neq 1 \in (f{}\mathcal{O}_X)_y $ and $0=(\mathcal{O}/\mathcal{K})_y$, how can we deduce contradiction..? – Plantation May 24 '22 at 00:34
  • And I wondering also that why 4 th $\Rightarrow$ is true. Let $\mathcal{F}$ be a sheaf on $Y$. Then $1 \in \mathcal{F}_y$ is defined as the class $[(V, 1)]$ for some neighborhodd $y \in V$ and $1 \in \mathcal{F}(V)$. And..although $ 0\neq 1 \in \mathcal{F}_y$, is there a possibility of existence of some neighborhood $ y\in V' \subset Y$ such that $\mathcal{F}(V')$ is zero?? – Plantation May 24 '22 at 01:53
  • O.K. For the first question, I think that I understand. Let $f : A \to B$ be a ring homomorphism. $f$ sends $0_A$ to $0_B$ and sends $1_A$ to $1_B$. If $B$ is not zero, then $0_B \neq 1_B$. And if $A$ is zero, then $0_A = 1_A$. But in this case it contradicts to that $f$ is a function! – Plantation May 24 '22 at 02:31
  • So remaining question is, why 4th $\Rightarrow$ is true ~And it seems that it can be proved by the similar argument as above. (?) Thank you. – Plantation May 24 '22 at 02:32
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    @Plantation That's very simple from definition of colimit: if two elements are equal somewhere, they are equal in the colimit. – Lao-tzu May 24 '22 at 07:52
  • O.K. I see. Thank you.~~ – Plantation May 24 '22 at 07:58
  • This is not true without additional conditions - see here for an example where the support of $\mathcal{O}_Y/\mathcal{K}$ is strictly greater than $\overline{f(X)}$. – KReiser May 24 '22 at 13:45
  • @KReiser Sorry, can you be precise about the "additional conditions" (maybe qc)? Maybe can you tell anything wrong in the above argument? – Lao-tzu May 24 '22 at 14:47
  • @Lao-tzu sorry, I thought I linked an answer where I explained the additional conditions. You need either quasi-compactness of the morphism or that the source is reduced - see some discussion here. I don't immediately see the problem in your argument, but it's also formatted in a way that is really difficult for me to read and I haven't had the time between my other commitments today to dissect it. – KReiser May 24 '22 at 17:14
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Here's one way to see this if we assume that $f$ is quasi-compact:

First note that the statement is true if both $X,Y$ are affine, say $X=\operatorname{Spec}(B)$ and $Y=\operatorname{Spec}(A)$. Indeed in this case, we have that $\overline{f(X)} = V(I)$, where $I$ is the kernel of the map $A\rightarrow B$. Of course here we have that $\operatorname{Supp}(A/I) = V(I)$.

Now let's suppose that $f$ is quasi-compact. We can assume that $Y$ is affine. Since $f$ is quasi-compact, we can cover $X$ by finitely many affine opens $U_i$, and we have that $\overline{f(X)} = \overline{ \bigcup_i f(U_i) } = \bigcup_i \overline{f(U_i)}$ (we used finiteness for the last equality).

By the above, we have that $\overline{f(U_i)} = \operatorname{Supp}(\mathcal{O}_Y/K_{U_i})$, where $K_{U_i}$ is the kernel of the map $\mathcal{O}_Y \rightarrow (f|_{U_i})_* \mathcal{O}_{U_i}$. We claim that $\operatorname{Supp}(\mathcal{O}_Y/K_{U_i}) \subseteq \operatorname{Supp}(\mathcal{O}_Y/K_X)$, where $K_X$ is the kernel of the map $\mathcal{O}_Y \rightarrow f_* \mathcal{O}_X$. This follows from the fact that $K_X \subseteq K_{U_i}$.

This implies that $\overline{f(X)} \subseteq \operatorname{Supp}(\mathcal{O}_Y/K_X)$.

loch
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  • Don't you need $Y$ compact for the reduction to $Y$ affine to work? – user347489 Jul 30 '18 at 22:41
  • @user347489 ah I think you're right – loch Jul 30 '18 at 22:50
  • I don't see why one needs $Y$ affine for the reduction to work - we have $\overline{f(X)}^X \cap U = \overline{U \cap f(X)}^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$.

    Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $\coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $\overline{f(X)}$, known as scheme-theoretic image.

     – Johann Haas Jul 31 '18 at 13:43
  • @JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer. – loch Jul 31 '18 at 13:58
  • @loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures. – Johann Haas Jul 31 '18 at 18:13
  • @JohannHaas My own answer isn't true for arbitrary locally ringed spaces I guess, at least one key observation—for the structure sheaf of a scheme $X\neq \emptyset$, $\mathcal{O}_X(U)\neq 0\Leftrightarrow U\neq \emptyset$—need not be true for arbitrary locally ringed spaces; since I think for a locally ringed spaces $X$, $\mathcal{O}_X(U)=0$ for some $U\neq \emptyset$ is allowed. – Lao-tzu Aug 01 '18 at 02:14
  • @Lao-tzu The zero ring does not have a unique maximal ideal, so it isn't a local ring. So the equivalence is actually true for all locally ringed spaces (but not for all ringed spaces) – Johann Haas Aug 02 '18 at 11:32
  • @Johann Haas That's quite good! – Lao-tzu Aug 02 '18 at 12:02
  • Can I ask question? Why we may assume that $Y$ is affine? In general, since $f$ is quasi-compact, let's cover $Y$ by affine open subsets ${ U_i }$ such that each $f^{-1}(U_i)$ are quasi-compact. Then $f|{f^{-1}(U_i)}^{U_i} : f^{-1}(U_i) \to U_i$ are also quasi-compact. Now, for each i, let $K_i :=\operatorname{ker}( f|{f^{-1}(U_i)}^{U_i})^{\flat} : \mathcal{O}Y|{U_i} \to (f|{f^{-1}(U_i)}^{U_i}){*}\mathcal{O}X|{f^{-1}(U_i)}$. Then by the reduction assumption, $\operatorname{Supp}(\mathcal{O}{U{i}}/K_i)=\overline{f(f^{-1}(U_i))}^{U_i}=\overline{f(X) \cap U_i}^{U_i}$. – Plantation May 25 '22 at 12:12
  • My question is, $\bigcup_i(\operatorname{Supp}(\mathcal{O}_{U_i}/K_i) = \operatorname{Supp}(\mathcal{O}_Y/K)$? – Plantation May 25 '22 at 12:14