# Question about the sheaf of differentials of a fibre

Let $$f: X \to Y$$ be a morphism of finite type of noetherian schemes. Let $$f(x) = y$$. I would like to see a proof of $$(\Omega_{ f^{-1}(y) / \operatorname{Spec} \kappa (y) })_x = (\Omega_{X/Y})_x \otimes_{O_{Y,y}} \kappa (y).$$ This is mentioned in the proof of Theorem 3 Section III. 5 in Mumford's Red Book without explanation.

I have tried to reduce to the affine case and apply facts about Kahler differentials on page 186 in Matsumura's Commutative Algebra, but I have not managed to make this work yet... Thank you.

Intuitively, both sides are taking the stalk at $$x$$ of the sheaf of relative differentials along the fiber of $$X_y\to \{y\}$$ of the map $$X\to Y$$ - one by restricting to the fiber direction then localizing at $$x$$, and the other by localizing at $$x$$ and then restricting to the fiber direction. The claim is that these procedures commute.

To prove this algebraically, our key ingredients are as follows:

1. $$\Omega_{X/Y}$$ is a quasi-coherent sheaf.
2. If we have maps of rings $$R\to R'$$ and $$R\to S$$, letting $$S'=S\otimes_R R'$$, then we have that $$\Omega_{S/R}\otimes_R R'=\Omega_{S'/R'}$$. (See Stacks 00RV for a refresher if you need it.)
3. If $$A\to B$$ is a ring map and $$S\subset A$$ is a multiplicative subset mapping to invertible elements of $$B$$, then $$\Omega_{B/A}=\Omega_{B/S^{-1}A}$$.
4. If $$A\to B$$ is a ring map and $$S\subset B$$ is a multiplicative subset, then $$S^{-1}\Omega_{B/A}=\Omega_{S^{-1}B/A}$$. (See Stacks 00RT for a refresher on 3 and 4 if you need it.)

By 1), we may reduce to the affine case: suppose $$X=\operatorname{Spec} B$$ and $$Y=\operatorname{Spec} A$$, $$f$$ corresponds to a ring map $$\varphi:A\to B$$, and $$x,y$$ correspond to prime ideals $$\mathfrak{q}\subset B,\mathfrak{p}\subset A$$ respectively with $$\varphi^{-1}(\mathfrak{q})=\mathfrak{p}$$. Then the fiber diagram

$$\require{AMScd} \begin{CD} X_y @>{}>> X\\ @VVV @VVV \\ \operatorname{Spec} k(y) @>{}>> Y \end{CD}$$

corresponds to the diagram of rings

$$\require{AMScd} \begin{CD} B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p} @<<< B\\ @AAA @AAA \\ k(y)=A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p} @<<< A \end{CD}$$

and $$\Omega_{X/Y}$$ is the $$\mathcal{O}_X$$-module associated to the $$B$$-module $$\Omega_{B/A}$$. Also by quasi-coherentness, we have that $$(\Omega_{X/Y})_x=(\Omega_{B/A})_\mathfrak{q}$$. As $$\mathcal{O}_{Y,y}=A_\mathfrak{p}$$ and $$k(y)=A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}$$, we see that the right hand side of your desired isomorphism is $$(\Omega_{B/A})_\mathfrak{q}\otimes_{A_\mathfrak{p}} A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}$$. By 4), we have that $$(\Omega_{B/A})_\mathfrak{q}= \Omega_{B_\mathfrak{q}/A}$$, and as all the elements in $$A\setminus \mathfrak{p}$$ map to elements in $$B\setminus \mathfrak{q}$$, we may apply 3) to see that $$\Omega_{B_\mathfrak{q}/A}=\Omega_{B_\mathfrak{q}/A_\mathfrak{p}}$$. Now applying 2), we see that $$(\Omega_{B/A})_\mathfrak{q}\otimes_{A_\mathfrak{p}} A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p} = \Omega_{B_\mathfrak{q}/A_\mathfrak{p}} \otimes_{A_\mathfrak{p}} A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p} = \Omega_{(B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q})/(A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p})}.$$

On the other hand, since $$f^{-1}(y)\to\operatorname{Spec} k(y)$$ is given by $$\operatorname{Spec} B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p}\to \operatorname{Spec} A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}$$, the left hand side of your desired isomorphism is $$(\Omega_{(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p})/(A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p})})_\mathfrak{q}$$ which is exactly $$\Omega_{(B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q})/(A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p})}$$ by 4). So we're done.