# Unramified map between local noetherian rings

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??

• It's true if $K/F$ is separable and to have unramifiedness at the generic point of $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ you need separability of the residue field extensions (which, because they are generic points, is just $K/F$). – Alex Youcis Sep 11 '19 at 16:47
• Having that $\Omega_{B/A}=0$ the map should be unramified (also etale in this particular case) so in particular unramified at the generic point no? – Tommaso Scognamiglio Sep 11 '19 at 16:51
• To check that $\Omega_{B/A}=0$ one just needs to do that at the closed point where the condition on the maximal ideals and on the residue field should be enough to conclude that the module of differentials is zero. – Tommaso Scognamiglio Sep 11 '19 at 16:54
• Why I'm saying to you is that it sounds like you're saying that the map $f:\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is unramified if $\mathfrak{m}B=\mathfrak{n}$ and $A/\mathfrak{m}\to B/\mathfrak{n}$ is an isomorphism. But, this is not true. Presumably you want to use the criterion "a map $X\to Y$ is etale if it's flat and unramified, and its urnamified if for all $x\in X$ with image $y\in Y$ we have that $\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$ and $k(x)/k(y)$ is a separable extension" But, you have only checked this at the closed point of $\mathrm{Spec}(B)$, not the generic point. – Alex Youcis Sep 11 '19 at 16:54
• To check that $\Omega^1_{B/A}$ is what on the closed point? – Alex Youcis Sep 11 '19 at 16:57

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.