Lemma 10.151.9. Let $k$ be a field. Let
be a finite type ring map. Then $\varphi $ is étale if and only if we have the following two conditions: (a) the local rings of $A$ at maximal ideals have dimension $n$, and (b) the elements $\text{d}(a_1), \ldots , \text{d}(a_ n)$ generate $\Omega _{A/k}$ as an $A$-module.
Proof. Assume (a) and (b). Condition (b) implies that $\Omega _{A/k[x_1, \ldots , x_ n]} = 0$ and hence $\varphi $ is unramified. Thus it suffices to prove that $\varphi $ is flat, see Lemma 10.151.8. Let $\mathfrak m \subset A$ be a maximal ideal. Set $X = \mathop{\mathrm{Spec}}(A)$ and denote $x \in X$ the closed point corresponding to $\mathfrak m$. Then $\dim (A_\mathfrak m)$ is $\dim _ x X$, see Lemma 10.114.6. Thus by Lemma 10.140.3 we see that if (a) and (b) hold, then $A_\mathfrak m$ is a regular local ring for every maximal ideal $\mathfrak m$. Then $k[x_1, \ldots , x_ n]_{\varphi ^{-1}(\mathfrak m)} \to A_\mathfrak m$ is flat by Lemma 10.128.1 (and the fact that a regular local ring is CM, see Lemma 10.106.3). Thus $\varphi $ is flat by Lemma 10.39.18.
Assume $\varphi $ is étale. Then $\Omega _{A/k[x_1, \ldots , x_ n]} = 0$ and hence (b) holds. On the other hand, étale ring maps are flat (Lemma 10.143.3) and quasi-finite (Lemma 10.143.6). Hence for every maximal ideal $\mathfrak m$ of $A$ we my apply Lemma 10.112.7 to $k[x_1, \ldots , x_ n]_{\varphi ^{-1}(\mathfrak m)} \to A_\mathfrak m$ to see that $\dim (A_\mathfrak m) = n$ and hence (a) holds. $\square$
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