Proposition IV.2.3 in Hartshorne

Question

(2.2) Here was referring to the exact sequence $$0\to f^*\Omega_Y\to\Omega_X\to \Omega_{X/Y}\to 0.$$ My concern is: as $\Omega_{X/Y}\cong\mathscr{O}_R$, I cannot see why tensoring with $\Omega_X^{-1}$ preserves $\mathscr{O}_R$ ONLY but not for the others.

Thanks in advance for answering.

Answer 1

So Prop 2.2.a) says that $\Omega_{X/Y} \simeq \mathcal O_R$ (since the morphism is separable).

Then the right hand side is after tensoring $\Omega_{X/Y} \otimes \Omega_X^{-1} \simeq \mathcal O_R \otimes \Omega_X^{-1}$.

Since $X$ is a curve, $\Omega_X^{-1}$ is a line bundle.

Now use the fact that if $\mathscr L$ is a line bundle and $\mathscr F$ has discrete support, then $\mathscr L \otimes \mathscr F \simeq \mathscr F$.

This is "obvious", because since $\mathscr F$ has discrete support, we only need to define the isomorphism locally, and $\mathscr L(U) \simeq k$ for small enough $U$.

Answer 2

I think it is a consequence of the three following points:

• If $\mathcal{F}$ has a discrete support then $\mathcal{F}\otimes\mathcal{G}$ too
• If $\mathcal{L}$ is an invertible sheaf and $P$ is a point $(\mathcal{F}\otimes\mathcal{L})_P=\mathcal{F}_P$
• For coherents sheaves, if $\mathrm{Supp}(\mathcal{F})=\mathrm{Supp}(\mathcal{G})=Z$ and $\mathcal{F}_{|Z}\simeq\mathcal{G}_{|Z}$ then $\mathcal{F}\simeq\mathcal{G}$

Answer 3

With regards to showing that $\mathcal{O}_R \cong \Omega_{X/Y}$, I add some algebraic content that complements the other two answers.

Let $M,N$ be two finitely generated $A$-modules over a Noetherian ring $A$. Suppose that $\operatorname{Supp}M =\operatorname{Supp}N = \mathcal{S} \subset \operatorname{Spec}A$. Suppose further that $\dim \mathcal{S} = 0$. Then we also have $\mathcal{S} = \operatorname{Ass}M = \operatorname{Ass}N$. Now suppose that $M_P \cong N_P$ for every $P \in \mathcal{S}$.

Consider the multiplicatively closed set $S = A \setminus \bigcup_{P \in \mathcal{S}} P$. Then the canonical maps $M \rightarrow M_S$ and $N \rightarrow N_S$ are isomorphisms of $A$-modules. Thus it is enough to show that $M_S \cong N_S$ as $A_S$-modules. We hence reduce to the case where $A$ is Artinian and $\mathcal{S} = \operatorname{Spec}A$. By the structure theorem of Artinian rings we have $A = \prod_{P \in \operatorname{Spec}A} A_P$. As modules over the product ring $A$, $M$ and $N$ are themselves the products of their localizations at every $P \in \operatorname{Spec}A$, that is $M = \prod_{P \in \operatorname{Spec}A} M_P$ and $N = \prod_{P \in \operatorname{Spec}A} N_P$. Now the isomorphisms $M_P \cong N_P$ immediately give an isomorphism $M \cong N$.

Finally, $\mathcal{O}_R \cong \Omega_{X/Y}$ because at every affine open $\operatorname{Spec}A$ the two sheaves are sheafifications of $A$-modules $M,N$ as above.