Lemma 31.22.10. Let

be a commutative diagram of morphisms of schemes. Assume that $Y \to S$ is syntomic, $X \to S$ smooth, and $i$ an immersion. Then $i$ is a regular immersion.

Lemma 31.22.10. Let

\[ \xymatrix{ Y \ar[rd] \ar[rr]_ i & & X \ar[ld] \\ & S } \]

be a commutative diagram of morphisms of schemes. Assume that $Y \to S$ is syntomic, $X \to S$ smooth, and $i$ an immersion. Then $i$ is a regular immersion.

**Proof.**
After replacing $X$ by an open neighbourhood of $i(Y)$ we may assume that $i$ is a closed immersion. Let $T = i(Y)$ be the corresponding closed subscheme of $X$. Since $T \cong Y$ the morphism $T \to S$ is flat and of finite presentation (Morphisms, Lemmas 29.30.6 and 29.30.7). Also a smooth morphism is flat and locally of finite presentation (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. As $X_ s$ is locally of finite type over a field, it is Noetherian (Morphisms, Lemma 29.15.6). Thus we can check that $T_ s \subset X_ s$ is a quasi-regular immersion at points, see Lemma 31.20.8. Take $t \in T_ s$. By Morphisms, Lemma 29.30.9 the local ring $\mathcal{O}_{T_ s, t}$ is a local complete intersection over $\kappa (s)$. The local ring $\mathcal{O}_{X_ s, t}$ is regular, see Algebra, Lemma 10.140.3. By Algebra, Lemma 10.135.7 we see that the kernel of the surjection $\mathcal{O}_{X_ s, t} \to \mathcal{O}_{T_ s, t}$ is generated by a regular sequence, which is what we had to show.
$\square$

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