Lemma 31.22.8. Let $f : X \to S$ be a smooth morphism of schemes. Let $\sigma : S \to X$ be a section of $f$. Then $\sigma $ is a regular immersion.

**Proof.**
By Schemes, Lemma 26.21.10 the morphism $\sigma $ is an immersion. After replacing $X$ by an open neighbourhood of $\sigma (S)$ we may assume that $\sigma $ is a closed immersion. Let $T = \sigma (S)$ be the corresponding closed subscheme of $X$. Since $T \to S$ is an isomorphism it is flat and of finite presentation. Also a smooth morphism is flat and locally of finite presentation, see Morphisms, Lemmas 29.34.9 and 29.34.8. Thus, according to Lemma 31.22.7, it suffices to show that $T_ s \subset X_ s$ is a quasi-regular closed subscheme. This follows immediately from Morphisms, Lemma 29.34.20 but we can also see it directly as follows. Let $k$ be a field and let $A$ be a smooth $k$-algebra. Let $\mathfrak m \subset A$ be a maximal ideal whose residue field is $k$. Then $\mathfrak m$ is generated by a quasi-regular sequence, possibly after replacing $A$ by $A_ g$ for some $g \in A$, $g \not\in \mathfrak m$. In Algebra, Lemma 10.140.3 we proved that $A_{\mathfrak m}$ is a regular local ring, hence $\mathfrak mA_{\mathfrak m}$ is generated by a regular sequence. This does indeed imply that $\mathfrak m$ is generated by a regular sequence (after replacing $A$ by $A_ g$ for some $g \in A$, $g \not\in \mathfrak m$), see Algebra, Lemma 10.68.6.
$\square$

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