# Condition on the stalks for a morphism $X \rightarrow S$ to be unramified

Let $$f: X \rightarrow S$$ be locally of finite presentation. Let $$x \in X, s = f(x)$$. Let's say that $$f$$ is unramified at $$x$$ if $$\Omega_{X/S,x} = 0$$. I have already shown this is equivalent to saying that the diagonal $$\Delta: X \rightarrow X \times_S X$$ is a local isomorphism at $$x$$.

I want to show that $$f$$ is unramified at $$x$$ if and only if $$\mathfrak m_x$$ is generated by $$\mathfrak m_s$$ and $$\kappa(s) \subset \kappa(x)$$ is finite separable. The condition that $$\mathfrak m_x$$ is generated by $$\mathfrak m_s$$ already implies that the extension $$\kappa(s) \subset \kappa(x)$$ is finite.

I think I understand most of the proof. But there is a gap in my understanding in each of the implications.

$$\Leftarrow$$: I already know a result that says that since $$\kappa(s) \subset \kappa(x)$$ is finite and separable, $$\Omega_{\kappa(x)/\kappa(s)} = 0$$. If $$X_s \rightarrow \operatorname{Spec} \kappa(s)$$ is the fiber of $$f$$ at $$s$$, then the residue field $$\kappa(x)$$ is the same whether we think of $$x$$ as belonging to $$X$$ or $$X_s$$. Letting $$p: X_s \rightarrow X$$ be the projection, we have

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

As explained in my previous question, we have

$$(\Omega_{X_s/\kappa(s)})_x = \Omega_{X/S,x} \otimes_{\mathcal O_{S,s}} \kappa(s)$$ And in the answer to the question, it is explained that

$$\Omega_{X/S,x} \otimes_{\mathcal O_{S,s}} \kappa(s) = \Omega_{\mathcal O_{X,x}/\mathcal O_{S,s}} \otimes_{\mathcal O_{S,s}} \kappa(s)= \Omega_{\mathcal O_{X,x} \otimes_{\mathcal O_{S,s}}\kappa(s)/\kappa(s)} =\Omega_{\kappa(x)/\kappa(s)}$$ where in this last equality, we are using the fact that $$\mathcal O_{X,x} \otimes_{\mathcal O_{S,s}}\kappa(s) = \kappa(x)$$, and this requires the hypothesis that $$\mathfrak m_s$$ generates $$\mathfrak m_x$$. Putting this all together, we get $$\Omega_{X/S,x} \otimes_{\mathcal O_{S,s}} \kappa(s) = 0$$. But I don't see how this implies $$\Omega_{X/S,x} = 0$$. I would want to use Nakayama's lemma, but the problem is that $$\Omega_{X/S,x}$$ need not be a finitely generated $$\mathcal O_{S,s}$$-module.

$$\Rightarrow$$: Since $$f$$ is unramified at $$x$$, the fibre $$X_s \rightarrow \operatorname{Spec} \kappa(s)$$ is also unramified at $$x$$. Letting $$k = \kappa(s)$$, we use the fact that the diagonal map $$X_s \rightarrow X_s \times_k X_s$$ is a local isomorphism at $$x$$. We may produce a suitably small affine open neighborhood $$U = \operatorname{Spec} A$$ of $$x$$ in $$X_s$$, such that $$A$$ is a finitely generated $$k$$-algebra, and the diagonal map $$U \rightarrow U \times_k U$$ is an open immersion.

It's not difficult to see from here that $$U$$ must be finite, and $$A$$ a finite product of finite separable extensions of $$k$$. If $$x$$ is the prime (automatically maximal) ideal $$\mathfrak m$$ of $$A$$, then $$\kappa(x)$$ will be one of these finite separable extensions of $$k$$.

So $$\kappa(x)$$ is a finite separable extension of $$\kappa(s)$$. But I still don't understand why it follows that $$\mathfrak m_s$$ must generate $$\mathfrak m_x$$.

• Like you said tensoring with $\otimes_{\mathcal{O}_{S,s}} k(s)$ is the same as tensoring with $\otimes_{\mathcal{O}_{X,x}} k(x)$ so if that is zero you can use Nakayama. – wsokursk Mar 3 at 4:30

For the first implication: $$0=\Omega^1_{X/S,x}\otimes_{\mathcal{O}_{S,s}} k(s)=\Omega^1_{X/S,x} \otimes_{\mathcal{O}_{X,x}} (\mathcal{O}_{X,x}\otimes_{\mathcal{O}_{S,s}} k(s))=\Omega^1_{X/S,x}\otimes_{\mathcal{O}_{X,x}} k(x),$$ where the last equality holds by assumption. Now $$\Omega^1_{X/S,x}$$ is finitely generated over $$\mathcal{O}_{X,x}$$ and you can use Nakayama's Lemma.
For the second implication, observe $$\mathfrak{m}_s\subset \mathcal{O}_{S,s}$$ generating $$\mathfrak{m}_x\subset \mathcal{O}_{X,x}$$ is equivalent to proving the analogous statement for the base-change $$X_s\to \operatorname{Spec}(k(s))$$. Indeed, if you have a morphism $$A\to B$$ and a prime $$x=\mathfrak{P}$$ of $$B$$ lying above a prime $$s=\mathfrak{p}$$ of $$A$$, then $$\mathcal{O}_{X_s,\mathfrak{P}}=B_{\mathfrak{P}}/\mathfrak{p}B_{\mathfrak{P}}$$ and $$k(s)=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$.
In other words, you can assume that $$S=\operatorname{Spec}(k(s))$$. By what you wrote, we may assume that the morphism of induced by a $$k=k(s)$$-algebra $$k\to \prod_{i=1}^n k_i$$ with each $$k_i$$ being a finite separable extension of $$k$$. Now simply observe that if you take the prime in $$\prod_{i=1}^n k_i$$ corresponding to $$k_j$$, then the localization at this prime is already $$k_j$$, in other words the maximal ideal in the localization is zero and hence generated by the maximal ideal of $$k$$, namely the zero ideal.
• So the stalk of the fibre $(X_s)_x$ is a field if and only if $\mathfrak m_s$ generates $\mathfrak m_x$? – D_S Mar 3 at 14:28