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I'm trying to do the final part of Exercise 3.11.d in chapter II of Hartshorne, i.e. that if $(Z,\mathcal{O}_Z)$ is a reduced scheme and $f:(Z,\mathcal{O}_Z)\to(X,\mathcal{O}_X)$ is a scheme morphism, then the scheme theoretic image of $f$ is given by the reduced induced closed subscheme structure on $\overline{f(Z)}$. With the help of this answer by KReiser, I was able to prove that the ideal sheaf $\ker f^{\#}$ defines a closed subscheme $(Y,\mathcal{O}_Y)$ of $X$ (with $Y\subseteq X$), and as $Z$ is reduced, $Y$ carries the reduced induced closed subscheme structure. However, to conclude the solution of the problem, we still need to show that $Y=\overline{f(Z)}$, which one easily reduces to showing that $\overline{f(Z)}\cap U=V(\ker f^{\#}(U))$ for all affine open $U\subseteq X$. And this is where I'm stuck. So how can one show this? Thanks in advance!

Redundant Aunt
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  • What's insufficient about the discussion from your previous question here? As described there, either the hypothesis that $f$ is quasi-compact or the hypothesis that $X$ is reduced will let you reach your desired conclusion. – KReiser Nov 07 '20 at 11:57
  • @KReiser You're right, this certainly helps, but for me there is still something missing to conclude. I edited the question to better highlight the problem. Thanks for pointing it out! – Redundant Aunt Nov 07 '20 at 16:03

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Let $U=\operatorname{Spec} A$ be an affine open, and cover $f^{-1}(U)$ by $\operatorname{Spec} B_i$ as $i\in I$ ranges over an index set. Then $\operatorname{Spec} (B_i)_g$ cover $f^{-1}(\operatorname{Spec} A_g)$. If $g\in\ker f^\sharp(U)$, then $(B_i)_g=0$ and so $f^{-1}(\operatorname{Spec} A_g)=\emptyset$ and $\ker f^\sharp(\operatorname{Spec} A_g)=A_g$, so the underlying set of the scheme-theoretic image is a subset of $\overline{f(Z)}$.

On the other hand, the underlying set of the scheme-theoretic image can't be smaller than $\overline{f(Z)}$: if it was, then there exists a point $z$ in $f(Z)$ not in the scheme-theoretic image, and a local function which vanishes on the image but not at $z$ defines a local function on $Z$ which is nonvanishing by pullback. So the underlying sets of $\overline{f(Z)}$ and $\operatorname{im}(f:Z\to X)$ are equal, and both are reduced (check that $\ker f^\sharp(U)$ is radical by the computations in the linked post). By the uniqueness of the reduced induced structure, they must be the same subscheme.

KReiser
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  • Thanks! Do I understand your argument correctly: if by contradiction $Y\subsetneq \overline{f(Z)}$, then $f(Z)\not\subseteq Y$, and thus there is a $z\in Z$ with $f(z)\notin Y$. But then there exists an affine neighborhood $U$ of $f(z)$ with $f(z)\in U\subseteq X\setminus Y$, and thus $\emptyset=U\cap Y=V(\ker f^{#}(U))$ which implies $f^{#}(U)=0$ and thus $\mathcal{O}_Z(f^{-1}U)$ is the trivial ring. Then by restriction also every $\mathcal{O}_Z(V)$ with $V\subseteq f^{-1}U$ is trivial, which implies that every affine open of $f^{-1}U$ is empty, so it must be empty itself, contradiction. – Redundant Aunt Nov 08 '20 at 10:33
  • Also, doesn't the argument in the beginning rather show that the underlying set of the scheme-theoretic image contains $\overline{f(Z)}$? Because from your argument we see that $f^{-1}(\operatorname{Spec} A_g)=\emptyset$, and thus $\overline{f(Z)}\cap\operatorname{Spec} A_g=\emptyset$ for all $g\in\ker f^{#}(U)$. This implies $\overline{f(Z)}\cap U\subseteq \bigcap_{g\in\ker f^{#}(U)}V_U(g)=Y\cap U$. – Redundant Aunt Nov 08 '20 at 11:08
  • @RedundantAunt your first comment seems a bit overwrought, but sure, that's one way to do it. I'd prefer saying $f:Z\to X$ factors through $i:Y\to X$ iff the composite $\mathcal{I}Y\to \mathcal{O}_X\to f*\mathcal{O}_Z$ is zero, so a function vanishing on the image but not $z$ demonstrates that this is not the case. Your second comment is a valid way to proceed - I guess I didn't see that earlier. Thanks for the alternate strategy. – KReiser Nov 08 '20 at 22:41
  • Maybe I didn’t make this clear enough in the second question: what I was asking is that to me, your first argument seems to prove that the scheme-theoretic image contains $\overline{f(Z)}$, rather than ‘is contained in’. Could you explain why your first argument shows $Y\subseteq \overline{f(Z)}$? And by the way, thanks for your repeated and very nice answers to my algebraic geometry questions! I’m currently learning as much as I can and planning to do a PhD afterwards, so having someone helping out with all these nitty-gritty details is very helpful :) – Redundant Aunt Nov 09 '20 at 06:44
  • @RedundantAunt Let $U=\operatorname{Spec} A$ be an affine open subset of $X$ as in the post. If $\operatorname{Spec} A_g$ doesn't contain $\overline{f(Z)}$, then $I_g=A_g$, and by the compatibility condition, this means $g^n\in I\subset A$ for some $n$. So every function vanishing on $\overline{f(Z)}$ is in the ideal cutting out the scheme-theoretic image inside $U$ up to radicals. By the inclusion-reversing bijection, this means that the underlying set of the scheme-theoretic image is contained in the set-theoretic image. (Apologies for the mixup - I must have read $\supset$ instead.) – KReiser Nov 09 '20 at 06:55