# If $\mathfrak m_s$ generates $\mathfrak m_x$ and $\kappa(x)/\kappa(s)$ is finite separable, then $\Omega_{X/S,x} = 0$

I'm trying to understand the various equivalent definitions of an unramified morphism of schemes. Let $$f: X \rightarrow S$$ be a morphism of schemes which is locally of finite type, $$x \in X$$, and $$s = f(x)$$. Let's say that $$f$$ is unramified if the stalk of $$\Omega_{X/S}$$ at $$x$$ is zero. I want to understand the following result:

If the maximal ideal $$\mathfrak m_x$$ is generated by the maximal ideal $$\mathcal m_s$$, and if $$\kappa(s) \subset \kappa(x)$$ is a finite separable extension, then $$f$$ is unramified at $$x$$.

Let $$k = \operatorname{Spec} \kappa(s)$$, $$X_s = X \times_S \operatorname{Spec} k$$ be the fiber of $$s$$, and $$p: X_s \rightarrow X$$ the "projection." We have

$$p^{\ast} (\Omega_{X/S}) = \Omega_{X_s/k}$$

And therefore

$$(\Omega_{X_s/k})_x = (\Omega_{X/S})_x \otimes_{\mathcal O_{X,x}} \mathcal O_{X_s,x} = (\Omega_{X/S})_x \otimes_{\mathcal O_{X,x}}(\mathcal O_{X,x} \otimes_{\mathcal O_{S,s}} \kappa(s)) = (\Omega_{X/S})_x \otimes_{\mathcal O_{S,s}} \kappa(s)$$

I'm trying to reduce the problem to the special case when $$S$$ is the spectrum of the field. I thought I could use Nakayama's lemma to say that if we show $$(\Omega_{X_s/k})_x = 0$$, then $$(\Omega_{X/S})_x = 0$$. But I do not know that $$(\mathcal O_{X/S})_x$$ is finitely generated as an $$\mathcal O_{S,s}$$-module. It is only finitely generated as an $$\mathcal O_{X,x}$$-module.

How should I go about proving the result? And where is the hypothesis that $$\mathfrak m_s$$ generates $$\mathfrak m_x$$ coming in?

Using Property (1) in your previous question, you can reduce your last term in the middle of the post the following case. Let me use the notation $$(S,n) \subset (T,m)$$, where $$m \cap S= n$$ and $$S/n \subset T/m$$ is finite and separable. Your last expression is then $$\Omega_{(T/S)} \otimes_S S/n \cong \Omega_{(T\otimes_S S/n)/(S/n)},$$ where the isomorphism follows from (1). Notice that $$T \otimes_S S/n \cong T/nT = T/m$$ since $$nT = m$$. As $$S/n \subset T/m$$ is finitely and separable, we see that the module in question is zero.
• Thank you. So this shows that $\Omega_{T/S} \otimes_S (S/n) = 0$. But how does this imply $\Omega_{T/S} = 0$? – D_S Mar 2 at 18:50
• You have it by the last equality after "and therefore". It implies $\Omega_{k(m)/k(n)} = 0$ because $(*) \otimes_S (S/n) = (*) \otimes_T T \otimes_S (S/n) = (*) \otimes_T (T/m)$ so you can use nakayama. – wsokursk Mar 3 at 4:14