Let $f:X\rightarrow Y$ be a morphism of schemes and let $\Omega_{X/Y}$ be the sheaf of relative differentials. Then, given a point $x\in X$, what can we say about $(\Omega_{X/Y})_x$?
In Hartshorne, Chapter 2 – Remark 8.9.2, he says:
The derivations $d:B\rightarrow \Omega_{B/A}$ glue together to give a map $d:\mathcal{O}_X\rightarrow \Omega_{X/Y}$ of sheaves of abelian groups on $X$, which is a derivation of the local rings at each point.
What I understood from the final line is $(\Omega_{X/Y})_x\cong \Omega_{\mathcal{O}_{X,x}/\mathcal{O}_{Y,f(x)}}$. But, this does not seem to make sense to me because if we take a morphism of rings $A\rightarrow B$ and a prime ideal $p\subset B$ then Hartshorne's Proposition 8.2A in the same chapter tells us that $(\Omega_{B/A})_p\cong \Omega_{B_p/A}$ and not $\Omega_{B_p/A_{q}}$, where $q=p^c$, the contraction of p.
Best Answer
Actually, $\Omega_{B_p/A}=\Omega_{B_p/A_q}$ in your case!
More generally, if $A\to B$ is a morphism of rings and $S\subset A$ is a multiplicatively closed subset of elements which all map to invertible elements, then $\Omega_{B/A}=\Omega_{B/S^{-1}A}$. We can prove this by looking at what happens to $1=\varphi(s)\varphi(s)^{-1}$ when taking $d$: $$d(1)=d(\varphi(s)\varphi(s)^{-1})$$ $$0=\varphi(s)d(\varphi(s)^{-1})+\varphi(s)^{-1}d(\varphi(s))$$ $$0=\varphi(s)d(\varphi(s)^{-1})$$ $$0=d(\varphi(s)^{-1})$$
where we've used the Leibniz rule, the fact that $d(\varphi(a))=0$ for any $a\in A$, and the fact that $\varphi(s)$ is invertible.