There is a lot of question about this subject but I can't get any comprehensive explanation for me.

In his section II.8 Hartshorne defines the sheaf of differential for $f:X\to Y$: he considers the diagonal morphism $\Delta:X\to X\times_Y X$ and says that it is a closed immersion in $W\subset X\times_Y X$. Telling $\mathscr{I}$ the ideal sheaf of $\Delta(X)$ in an open $W$ (union of the $U\times_Y U$ with the $U$ affines) he defines the sheaf of differentials $\Omega$ to be $\Delta^*(\mathscr{I}/\mathscr{I}^2)$.

My question is: why can we consider $\mathscr{I}/\mathscr{I}^2$ as a $\mathcal{O}_{X\times_Y X}$-module? it is a $\mathcal{O}_W$-module (ideal in fact), it is the kernel of $j^*:\mathcal{O}_W\to j_*\mathcal{O}_X$ with $j:X\to W$ induced by $\Delta$.

Should we consider in fact $i_*(\mathscr{I}/\mathscr{I}^2)$ where $i:W\to X\times_Y X$ is the canonical open immersion?

  • $\begingroup$ Is "why can we consider $J$ as an $\mathcal{O}_{X\times_Y X}$-module" a typo or your actual question? The error in this sentence is that we intend to consider $J/J^2$, not $J$. $\endgroup$
    – KReiser
    Commented Jan 6, 2020 at 23:44
  • $\begingroup$ Yes I mean $\mathscr{I}/\mathscr{I}^2$. I correct the question. $\endgroup$ Commented Jan 7, 2020 at 7:08
  • $\begingroup$ Rereading your question, I must admit I'm more confused right now. $J/J^2$ is a perfectly good $\mathcal{O}_{X\times_Y X}$ module: on any scheme $Z$, any ideal sheaf $\mathcal{I}\subset\mathcal{O}_Z$ is a perfectly good $\mathcal{O}_Z$ module, and for any $\mathcal{F}$ which is an $\mathcal{O}_Z$ module, we can take the quotient $\mathcal{F/IF}$, which is naturally a $\mathcal{O}_Z$ module. Applying this to the situation at hand, we can consider $J/J^2$ as a $\mathcal{O}_{X\times_YZ}$ module. Is this actually what you're having issues with? Where do you encounter difficulty in this reasoning? $\endgroup$
    – KReiser
    Commented Jan 7, 2020 at 7:44
  • $\begingroup$ By definition $\mathscr{I}$ is the ideal sheaf of the closed immersion $j:X\to W$ so $\mathscr{I}$ is a $\mathcal{O}_W$-module and not a $\mathcal{O}_{X\times_Y X}$-module. It is my problem: why can we switch $W$ and $X\times_Y X$? $\endgroup$ Commented Jan 7, 2020 at 8:12

1 Answer 1


Let's start by reproducing the text.

Let $f:X\to Y$ be a morphism of schemes. We consider the diagonal morphism $\Delta: X\to X\times_YX$. It follows from the proof of (4.2) that $\Delta$ gives an isomorphism of $X$ onto its image $\Delta(X)$, which is a locally closed subscheme of $X\times_YX$, i.e., a closed subscheme of an open subset $W\subset X\times_YX$.

Definition. Let $\mathscr{I}$ be the sheaf of ideals of $\Delta(X)$ in $W$. Then we can define the sheaf of relative differentials of $X$ over $Y$ to be the sheaf $\Omega_{X/Y}=\Delta^*(\mathscr{I/I}^2)$ on $X$.

Hartshorne is not claiming that $\mathscr{J/J}^2$ is a sheaf on $X\times_Y X$. Instead, he's using $\Delta$ to refer to the map which is the same as $\Delta:X\to X\times_YX$ but with the codomain restricted to to $W$ (which we can talk about since $\Delta$ lands in $W$ by definition of $W$).

  • $\begingroup$ Thanks so much for the time you devote to me! So, if I understand, in my notation the very definition of Hartshorne is $\Omega_{X/Y}=j^*(\mathscr{I}/\mathscr{I}^2)$. In fact he assimilates $\Delta$ and its canonical factorization trough $W$. $\endgroup$ Commented Jan 7, 2020 at 8:34
  • $\begingroup$ Yes, that looks correct from your definition of $j$. One slightly annoying thing here (and what I would guess as the reason Hartshorne skipped over this) is that there's not really a standard convenient way to denote restricting the codomain of a map. $\endgroup$
    – KReiser
    Commented Jan 7, 2020 at 8:40

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