In order to apply II.8.9, we need $\dim_kM\otimes_{A}k=\dim_KM\otimes_AK$ to show $M$ is free, where $A$ is a local integral domain, but by the result of generic and closed points, we only have $\dim_k(\Omega_{X/Y})_x\otimes_{\mathcal{O}_{X,x}}k=\dim_{K(X)}(\Omega_{X/Y})_{\xi}\otimes_{\mathcal{O}_{X,\xi}}K(X)$, where $x$ is the closed point and $\xi$ is the generic point (thus $k(x)=k,k(\xi)=K(X)$ and $\Omega_{X/Y}\otimes k(x)=(\Omega_{X/Y})_x\otimes_{\mathcal{O}_{X,x}} k$).
So we don't have the same local ring $A$ and $(\Omega_{X/Y})_x\not=(\Omega_{X/Y})_\xi$ in general. Moreover, I think we want $M$ to be the $\Omega_{X/Y}(U)$ instead of the stalk. I wonder how to apply the II.8.9 correctly?
Best Answer
Let us recall the text of Lemma II.8.9:
A couple reminders: we have the exact sequence because the functor $-\otimes_A K$ is exact, because it's the same as localizing at $(0)$ and localization is exact. $R$ is torsion-free because any torsion would have to be $\mathfrak{m}$-torsion, but this would be killed by tensoring with $k$, implying that $\dim_k M\otimes_A k < r$.
To apply this in our situation, we immediately notice that we may reduce to the case of $X$ affine integral, say $X=\operatorname{Spec} R$. Then $\Omega_{X/Y}$, being a coherent sheaf on $X$, is of the form $\widetilde{F}$ for a finitely-generated $R$-module $F$. Now for a closed point $x\in X$ and the generic point $\zeta\in X$, $(\Omega_{X/Y})_x=F_x$ and $(\Omega_{X/Y})_\zeta = F_{(0)}$ by the properties of stalks of quasi-coherent sheaves on affine schemes. This puts us in the situation of II.8.9 by taking $A=R_x$, $M=F_x$, $k=k(x)$, $K=Frac(R)$, and $r=n$. The claim that $\dim_{k(x)} \Omega_{X/Y}\otimes k(x)=n$ translates to $\dim_k M\otimes_A k=r$, so we get a surjection $A^r\to M$ and thus an exact sequence of the form $$0\to R \to A^r\to M\to 0$$ as in the proof. Tensoring with $K$ (aka localizing at $(0)$) preserves exactness, and the middle term becomes $K^r$, which implies $\dim_K M\otimes_A K\leq r$. On the other hand, the statement that $\dim_{k(\zeta)} \Omega_{X/Y}\otimes k(\zeta) \geq n$ is equivalent to $\dim_K M\otimes_A K\geq r$, which means that $\dim_K M\otimes_A K= r$. Now we can apply the conclusion of II.8.9 to see that $(\Omega_{X/Y})_x$ is free of rank $n$.
By exercise II.5.7(a), cited in your previous question, this implies that $\Omega_{X/Y}$ is locally free of rank $n$, and we are done.
The two key ideas in this proof are some foundational results on stalks for quasicoherent sheaves over affines, and some algebra with local rings. Your recent questions have shown some lack of understanding of the first topic - I would recommend you take another look at what's going on there, because these statements are necessary background knowledge to get stuff done in algebraic geometry.