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.
 A: 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:

*

*$\Omega_{X/Y}$ is a quasi-coherent sheaf.

*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.)

*If $A\to B$ is a ring map and $S\subset A$ is a multiplicative subset mapping to invertible elements of $B$, then $\Omega_{B/A}=\Omega_{B/S^{-1}A}$.

*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.
