I have a question that may be trivial, but I can not figure it out. It concerns Hartshorne's following remark about smooth morphisms:
If $X$ and Y are schemes of finite type over a field $k$, with X being integral, and if $f:X\rightarrow Y$ is a smooth morphism of relative dimension $n$, then the third condition of the definition is equivalent to saying that $\Omega _{X/Y}$ is locally free on $X$ of rank $n$.
The third condition he is referring to is
For each point $x\in X$, $dim_{k(x)}(\Omega _{X/Y}\otimes k(x))=n$ $\quad$ (ie $dim_{k(x)}((\Omega _{X/Y})_{x}\otimes_{O_{x,X}} k(x))=n$)
To see this, we are supposed to used to use the following lemma
Let $A$ be a noetherian local integral domain with residue field $k$ and quotient field $K$. If $M$ is a finitely generated $A$-module and if $dim_k(M\otimes_A k)=dim_K(M\otimes_A K)=r$, then $M$ is free of rank $r$.
I understand that we are supposed to use this with $A=O_{x,X}$ for any point $x\in X$. $(\Omega _{X/Y})_{x} = \Omega _{O_{x,X}/O_{f(x),Y}}$ is indeed an $O_{x,X}$-module of finite type. However, I don't see how we can show that $dim_{K}(\Omega _{X/Y}\otimes K)=n$, where $K$ is now the quotient field of $O_{x,X}$.
Could someone please provide me some help ? I thank you in advance.