The ideal sheaf of the base change of a closed immersion

algebraic-geometryschemes

Let $i:Z\to X$ be a closed immersion of schemes of quasi-coherent ideal sheaf $\mathcal{I}$ and $f:Y\to X$ be a morphism of schemes.
$\require{AMScd}$
\begin{CD}
W @>{j}>> Y\\
@V g V V @VV f V\\
Z @>{i}>> X
\end{CD}

Let $W$ be the fiber product. Then $j:W\to Y$ is a closed immersion by [https://stacks.math.columbia.edu/tag/01QR]. What is the ideal sheaf $\mathcal{J}$ corresponding to $j$?

Some examples (maybe I am wrong):

1.If $f$ is affine, $Y=\underline{\mathrm{Spec}}(\mathcal{A})$, $\mathcal{A}$ being a quasi-coherent $O_X$-algebra. Then $\mathcal{I}\mathcal{A}$ is an $\mathcal{A}$-module and $\tilde{\mathcal{IA}}=\mathcal{J}$.

2.If $Y=\underline{\mathrm{Proj}}(\mathcal{R})$, $\mathcal{R}$ being a quasi-coherent graded $O_X$-algebra. Then $\tilde{\mathcal{IR}}=\mathcal{J}$.

3.If $f$ is flat, then by EGA3, Ch.III, Proposition 1.4.15, $j_*O_W=f^*i_*O_Z$. Recall the SES $$0\to\mathcal{I}\to O_X\to i_*O_Z\to0.$$
As $f$ is flat, we have a SES $$0\to f^*I\to O_Y\to j_*O_W\to0.$$ So, $\mathcal{J}=f^*I$

What about the general case?

Best Answer

If we follow the chain of lemmas cited in the proof of the lemma you linked, we arrive at Stacks Project, Lemma 26.4.7 (tag 01HQ), which tells us that, in general, $\mathcal{J} = \operatorname{Im}(f^* \mathcal{I} \to \mathcal{O}_Y)$.

The key point (which implies the universal property of the fiber product for the closed subscheme corresponding to $\mathcal{J}$) is that a morphism $\varphi\colon W \to X$ factors through the closed immersion $i\colon Z \to X$ if and only if the map $\varphi^* \mathcal{I} \to \varphi^* \mathcal{O}_X = \mathcal{O}_W$ is zero. The forward implication is trivial. For the reverse implication, the topological part of this statement can be checked on stalks. The statement on sheaves follows from the adjunction between $f^*$ and $f_*$, which gives a correspondence between maps $\varphi^* \mathcal{I} \to \mathcal{O}_W$ and maps $\mathcal{I} \to \varphi_* \mathcal{O}_W$, so since these maps are zero, we get a map $\mathcal{O}_X/\mathcal{I} \to \varphi_* \mathcal{O}_W$ and thus a factorization $W \to Z \to X$. Details can be found in Stacks Project, Lemma 26.4.6 (tag 01HP).

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