Lemma 29.35.14. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. Set $s = f(x)$. Assume $f$ is locally of finite type (resp. locally of finite presentation). The following are equivalent:

1. The morphism $f$ is unramified (resp. G-unramified) at $x$.

2. The fibre $X_ s$ is unramified over $\kappa (s)$ at $x$.

3. The $\mathcal{O}_{X, x}$-module $\Omega _{X/S, x}$ is zero.

4. The $\mathcal{O}_{X_ s, x}$-module $\Omega _{X_ s/s, x}$ is zero.

5. The $\kappa (x)$-vector space

$\Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x) = \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$

is zero.

6. We have $\mathfrak m_ s\mathcal{O}_{X, x} = \mathfrak m_ x$ and the field extension $\kappa (x)/\kappa (s)$ is finite separable.

Proof. Note that if $f$ is unramified at $x$, then we see that $\Omega _{X/S} = 0$ in a neighbourhood of $x$ by the definitions and the results on modules of differentials in Section 29.32. Hence (1) implies (3) and the vanishing of the right hand vector space in (5). It also implies (2) because by Lemma 29.32.10 the module of differentials $\Omega _{X_ s/s}$ of the fibre $X_ s$ over $\kappa (s)$ is the pullback of the module of differentials $\Omega _{X/S}$ of $X$ over $S$. This fact on modules of differentials also implies the displayed equality of vector spaces in part (4). By Lemma 29.32.12 the modules $\Omega _{X/S, x}$ and $\Omega _{X_ s/s, x}$ are of finite type. Hence the modules $\Omega _{X/S, x}$ and $\Omega _{X_ s/s, x}$ are zero if and only if the corresponding $\kappa (x)$-vector space in (4) is zero by Nakayama's Lemma (Algebra, Lemma 10.20.1). This in particular shows that (3), (4) and (5) are equivalent. The support of $\Omega _{X/S}$ is closed in $X$, see Modules, Lemma 17.9.6. Assumption (3) implies that $x$ is not in the support. Hence $\Omega _{X/S}$ is zero in a neighbourhood of $x$, which implies (1). The equivalence of (1) and (3) applied to $X_ s \to s$ implies the equivalence of (2) and (4). At this point we have seen that (1) – (5) are equivalent.

Alternatively you can use Algebra, Lemma 10.151.3 to see the equivalence of (1) – (5) more directly.

The equivalence of (1) and (6) follows from Lemma 29.35.12. It also follows more directly from Algebra, Lemmas 10.151.5 and 10.151.7. $\square$

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