# Definition of the sheaf of differential

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?

• 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 $J$. Commented Jan 6, 2020 at 23:44
• Yes I mean $\mathscr{I}/\mathscr{I}^2$. I correct the question. Commented Jan 7, 2020 at 7:08
• 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? Commented Jan 7, 2020 at 7:44
• 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 $X\times_Y X$? Commented Jan 7, 2020 at 8:12

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 $$W$$).
• 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 $W$. Commented Jan 7, 2020 at 8:34
• 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. Commented Jan 7, 2020 at 8:40