Unramified map between local noetherian rings

Question

Let $A \subseteq B$ a finite type inclusion of two regular local noetherian rings. Let's call $\mathfrak{m}, \mathfrak{n}$ the maximal ideals of $A,B$ respectively. Let's suppose that $\mathfrak{m}B=\mathfrak{n}$ and $\dfrac{A}{\mathfrak{m}} \cong \dfrac{B}{\mathfrak{n}}$.

With this condition, I've got that the map is unramified and so that $\Omega_{B/A}=0$. Now, I've got a doubt: given that, let's call $F,K$ the fraction fields respectively of $A,B$, one should get $\Omega_{K/F}=0$ which in general is not true.

What am I missing??

Answer 1

Claim: If $A\subset B$ is an inclusion of domains with $Frac(A)=F$ and $Frac(B)=K$ and $\Omega_{B/A}=0$, then $\Omega_{K/F}=0$ as well.

Proof: We use Stacks 00RT:

Let $\phi: A\to B$ be a ring map.

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

2. If $S\subset B$ is a multiplcative subset, then $S^{-1}\Omega_{B/A}=\Omega_{S^{-1}B/A}$.

First, letting $S=B\setminus 0$, we see that $$\Omega_{K/A}=\Omega_{S^{-1}B/A}=S^{-1}\Omega_{B/A}=S^{-1}0=0$$ by (2). But then with $S^{-1}=A\setminus 0$ and considering the composite $A\to B\to K$ we have $$\Omega_{K/F}=\Omega_{K/S^{-1}A}=\Omega_{K/A}$$ by (1). So this module $\Omega_{K/F}$ is in fact zero. $\blacksquare$

This applies to your situation and shows your (un)desired result. I think Alex Youcis' interpetation in the comments is a good one: the unramified locus of a morphism is open, and the only open subset of the spectrum of a regular local ring which contains the closed point is the whole spectrum. So if you have that your map is unramified at the closed point, it must be unramified at all points of the spectrum.