Lemma 58.13.4. Let $X$ be an integral normal scheme with function field $K$. Let $L/K$ be a finite extension. Let $Y \to X$ be the normalization of $X$ in $L$. The following are equivalent

$X$ is unramified in $L$ as defined in Section 58.11,

$Y \to X$ is an unramified morphism of schemes,

$Y \to X$ is an étale morphism of schemes,

$Y \to X$ is a finite étale morphism of schemes,

for $x \in X$ the projection $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is unramified,

same as in (5) but with $\mathcal{O}_{X, x}^ h$,

same as in (5) but with $\mathcal{O}_{X, x}^{sh}$,

for $x \in X$ the scheme theoretic fibre $Y_ x$ is étale over $x$ of degree $\geq [L : K]$.

If $L/K$ is Galois with Galois group $G$, then these are also equivalent to

for $y \in Y$ the group $I_ y = \{ g \in G \mid g(y) = y\text{ and } g \bmod \mathfrak m_ y = \text{id}_{\kappa (y)}\} $ is trivial.

**Proof.**
The equivalence of (1) and (2) is the definition of (1). The equivalence of (2), (3), and (4) is Lemma 58.11.1. It is straightforward to prove that (4) $\Rightarrow $ (5), (5) $\Rightarrow $ (6), (6) $\Rightarrow $ (7).

Assume (7). Observe that $\mathcal{O}_{X, x}^{sh}$ is a normal local domain (More on Algebra, Lemma 15.45.6). Let $L^{sh} = L \otimes _ K K_ x^{sh}$ where $K_ x^{sh}$ is the fraction field of $\mathcal{O}_{X, x}^{sh}$. Then $L^{sh} = \prod _{i = 1, \ldots , n} L_ i$ with $L_ i/K_ x^{sh}$ finite separable. By Algebra, Lemma 10.147.4 (and a limit argument we omit) we see that $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ is the integral closure of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ in $L^{sh}$. Hence by Lemma 58.11.1 (applied to the factors $L_ i$ of $L^{sh}$) we see that $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ is finite étale. Looking at the generic point we see that the degree is equal to $[L : K]$ and hence we see that (8) is true.

Assume (8). Assume that $x \in X$ and that the scheme theoretic fibre $Y_ x$ is étale over $x$ of degree $\geq [L : K]$. Observe that this means that $Y$ has $\geq [L : K]$ geometric points lying over $x$. We will show that $Y \to X$ is finite étale over a neighbourhood of $x$. This will prove (1) holds. To prove this we may assume $X = \mathop{\mathrm{Spec}}(R)$, the point $x$ corresponds to the prime $\mathfrak p \subset R$, and $Y = \mathop{\mathrm{Spec}}(S)$. We apply More on Morphisms, Lemma 37.42.1 and we find an étale neighbourhood $(U, u) \to (X, x)$ such that $Y \times _ X U = V_1 \amalg \ldots \amalg V_ m$ such that $V_ i$ has a unique point $v_ i$ lying over $u$ with $\kappa (v_ i)/\kappa (u)$ purely inseparable. Shrinking $U$ if necessary we may assume $U$ is a normal integral scheme with generic point $\xi $ (use Descent, Lemmas 35.16.3 and 35.18.2 and Properties, Lemma 28.7.5). By our remark on geometric points we see that $m \geq [L : K]$. On the other hand, by More on Morphisms, Lemma 37.19.2 we see that $\coprod V_ i \to U$ is the normalization of $U$ in $\mathop{\mathrm{Spec}}(L) \times _ X U$. As $K \subset \kappa (\xi )$ is finite separable, we can write $\mathop{\mathrm{Spec}}(L) \times _ X U = \mathop{\mathrm{Spec}}(\prod _{i = 1, \ldots , n} L_ i)$ with $L_ i/\kappa (\xi )$ finite and $[L : K] = \sum [L_ i : \kappa (\xi )]$. Since $V_ j$ is nonempty for each $j$ and $m \geq [L : K]$ we conclude that $m = n$ and $[L_ i : \kappa (\xi )] = 1$ for all $i$. Then $V_ j \to U$ is an isomorphism in particular étale, hence $Y \times _ X U \to U$ is étale. By Descent, Lemma 35.23.29 we conclude that $Y \to X$ is étale over the image of $U \to X$ (an open neighbourhood of $x$).

Assume $L/K$ is Galois and (9) holds. Then $Y \to X$ is étale by Lemma 58.12.5. We omit the proof that (1) implies (9).
$\square$

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