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Let $X\to S$ be a scheme.

Definition: A relative effective Cartier divisor on $X/S$ is a closed subscheme $D\subset X$ such that the ideal sheaf $I$ of $D$ is invertible and $D\to S$ is flat.

Let now $T\to S$ be another scheme over $S$ and denote by $X_T$ the fibered product $X\times_S T$. I read on a paper the following:

Claim: Let $D$ be a relative effective Cartier divisor on $X_T/T$ and $p:T'\to T$ be an arbitrary $S$-map of schemes. Then the pullback $p^*_{X_T}(I)$ is the ideal sheaf of the $T'$-flat closed subscheme $D_{T'}\subset X_{T'}$. Hence $D_{T'}$ is a relative effective Cartier divisor on $X_{T'}/T'$.

Can you please help me proving the claim? In particular, the author says:

Since $D$ is $T$-flat, $p^*_{X_T} (I)$ is equal to the ideal of $D_{T'}$

And I don't understand why.

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    $\begingroup$ This is exactly what flatness means, perhaps you should write down the pullbacks and ideal sheaves in the affine case to see this. $\endgroup$ Nov 23, 2013 at 22:36
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    $\begingroup$ @MartinBrandenburg: Dear Martin, I posted an attempt to prove the claim, but I still have doubts. Could you please help me clarifying? Thanks a lot, your help is precious as diamonds! $\endgroup$
    – Abramo
    Nov 25, 2013 at 9:25

1 Answer 1

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Ok I think I really got it now! A nice way to prove the above claim is to use a basic result about purity:

The functor $\square\otimes_A B$ is exact if and only if every short exact sequence of $A$-modules ending with $B$ is pure

Recall that a short exact sequence of $A$-modules is pure if it stays exact when tensored with any $A$-module. $\newcommand{\Spec}{\operatorname{Spec}}$ $\newcommand{\SES}[3]{0\to #1 \to #2 \to #3 \to 0}$

Let's work affine locally, with $$ T=\Spec(R), \quad T'=\Spec(R'), \quad X_T=\Spec(S), \quad X_{T'}=\Spec(S') $$ and let $I\subset X_T$ be the ideal of $D$ and $I'$ the ideal of $D'=p^*{D}$.

Flatness of the map $D\to T$ is equivalent to the functor $\square\otimes_R S/I$ being exact. By the above result this implies that every short exact sequence of $R$-modules ending with $S/I$ is pure. In particular the sequence of $D$ $$ \SES{I}{S}{S/I} $$ is pure. Hence tensoring with any $R$-module leaves it exact and in particular $$ \SES{I\otimes_R R'}{S\otimes_R R'}{S/I\otimes_R R'} $$ is exact. Since the above sequence can be rewritten as $$ \SES{p^*I}{S'}{S'/I'} $$ we deduce that $p^*I = I'$, as desired.

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    $\begingroup$ $\pi$ and its restriction toare flat $D'$ aren't necessarily going to be flat unless $T'\to T$ is. The point is that $\mathcal{O}_D$ is $T$-flat. Reduce to the affine case $T = \mathop{\text{Spec}}R$, $T' = \mathop{\text{Spec}} R'$, and $X = \mathop{\text{Spec}}S$, where exactness of $0\to I\to S\to S/I\to 0$ and flatness of $S/I\to R$ imply exactness of $0\to I\otimes_R R'\to S\otimes_R R'\to (S/I)\otimes_R R'\to 0$. $\endgroup$ Nov 27, 2013 at 1:14
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    $\begingroup$ Or look at EGA IV${}_2$ (2.1.8)(i) (which is the same argument, except in French) $\endgroup$ Nov 27, 2013 at 1:41

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