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 $\mathfrak m_x$.
Best Answer
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.