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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