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Background:

  • Let $X: \textbf{CRing} \to \textbf{Set}$ be a presheaf on the category of affine schemes and $Z \subseteq X$ a subfunctor. One defines $Z$ to be closed if for every ring $A$ and every morphism $f: \text{Hom}(A , -) \to X$ the inverse image $f^{-1}(Z)$ is of the form $R \mapsto \{ \varphi : A \to R | \varphi(I) = 0 \}$ for some ideal $I \subseteq A$.
  • The intersection of subfunctors is defined naively, as is the closure (denoted by $\overline{Z}$) of a subfunctor $Z \subseteq X$ (it is the intersection of all closed subfunctors of $X$ containing $Z$).
  • If $Y$ is another presheaf, the product of $X$ and $Y$ is also defined naively.

Context: In section 1.14 of Jens Jantzen's great book "Representations of Algebraic Groups", the following is stated: If $X$ and $Y$ are presheaves which are schemes over a noetherian ring $k$ and $Z \subseteq X$ is a subscheme, and if $Z, X$ are algebraic and $Y$ is flat, then $\overline{Z \times Y} = \overline{Z} \times Y$. For the proof, he references Demazure-Gabriel I, section 2, 4.14 (although in my copy of Bell's translation this reference unfortunately doesn't exist).

Actual Question: Is this true for general presheaves? I.e. if $X$ and $Y$ are presheaves and $Z \subseteq X$ is a subfunctor, is it true that $\overline{Z \times Y} = \overline{Z} \times Y$? I worry that it isn't because of the conditions in Jantzen stated above, but I haven't been able to decide either way. (Also side question: does anyone know the correct reference in the translation?)

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This is not true even for affine schemes. Let $k = \mathbb{Z}$, let $X = \operatorname{Spec} \mathbb{Z}$, let $Y = \operatorname{Spec} \mathbb{F}_p$, and let $Z \cong \operatorname{Spec} \mathbb{Z} [ p^{-1}]$. The closure of $Z$ in $X$ is $X$ itself, but $Z \times Y \cong \operatorname{Spec} \{ 0 \}$, which is already closed in $X \times Y$. Of course, $Y$ is not flat.

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    $\begingroup$ It's good to see you around, Zhen Lin! $\endgroup$
    – Tim Campion
    Commented Jul 17, 2020 at 21:16
  • $\begingroup$ Thanks so much for your answer!! (Even though I was hoping the opposite would be true ;). ) This is a great example! I'm actually secretly interested in the special case when $G$ is a $k$-group functor acting on a $k$-functor $X$. If $Y \subseteq X$ is a subfunctor and the action of $G$ on $X$ restricts to an action of $G$ on $Y$, do you happen to know if $G$ also acts on the closure $\overline{Y} \subseteq X$? $\endgroup$
    – Nate Gallup
    Commented Jul 23, 2020 at 0:10
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    $\begingroup$ I think you want $G$ to be at least flat over $k$ there. The analogous fact for continuous group actions has an abstract nonsense proof using the fact that a group action $G \times X \to X$ is automatically an open map and the fact that pullback along open maps preserves denseness. But for schemes even the trivial action $G \times 1 \to 1$ might not be open. $\endgroup$
    – Zhen Lin
    Commented Jul 23, 2020 at 7:51
  • $\begingroup$ Thanks again for your comment! This has been extremely helpful. If you don't mind my bombarding you with a few more questions... What do you mean by pullback along the group action preserves denseness? Just that the inverse image of $Y$ under $G \times X \to X$ (which is just $G \times Y$) is dense in the inverse image of $\overline{Y}$? $\endgroup$
    – Nate Gallup
    Commented Jul 31, 2020 at 19:57
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    $\begingroup$ Yes, precisely. Of course, the preimage of $\overline{Y}$ is closed in $G \times X$, so the preimage of $Y$ being a dense subset means that the closure is the preimage of $\overline{Y}$, as desired. Incidentally, I think it is misleading to think of the preimage of $Y$ by the group action as $G \times Y$ – although it is true when $Y$ is closed under the group action, if it is not, there will be twisting by the group action. $\endgroup$
    – Zhen Lin
    Commented Jul 31, 2020 at 23:32

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