Lemma 15.33.5. Let $R \to S$ be a ring map. The following are equivalent

1. $R \to S$ is syntomic (Algebra, Definition 10.136.1), and

2. $R \to S$ is flat and a local complete intersection.

Proof. Assume (1). Then $R \to S$ is flat by definition. By Algebra, Lemma 10.136.15 and Lemma 15.33.3 we see that it suffices to show a relative global complete intersection is a local complete intersection homomorphism which is Lemma 15.33.4.

Assume (2). A local complete intersection is of finite presentation because a Koszul-regular ideal is finitely generated. Let $R \to k$ be a map to a field. It suffices to show that $S' = S \otimes _ R k$ is a local complete intersection over $k$, see Algebra, Definition 10.135.1. Choose a prime $\mathfrak q' \subset S'$. Write $S = R[x_1, \ldots , x_ n]/I$. Then $S' = k[x_1, \ldots , x_ n]/I'$ where $I' \subset k[x_1, \ldots , x_ n]$ is the image of $I$. Let $\mathfrak p' \subset k[x_1, \ldots , x_ n]$, $\mathfrak q \subset S$, and $\mathfrak p \subset R[x_1, \ldots , x_ n]$ be the corresponding primes. By Definition 15.32.1 exists an $g \in R[x_1, \ldots , x_ n]$, $g \not\in \mathfrak p$ and $f_1, \ldots , f_ r \in R[x_1, \ldots , x_ n]_ g$ which form a Koszul-regular sequence generating $I_ g$. Since $S$ and hence $S_ g$ is flat over $R$ we see that the images $f'_1, \ldots , f'_ r$ in $k[x_1, \ldots , x_ n]_ g$ form a $H_1$-regular sequence generating $I'_ g$, see Lemma 15.31.4. Thus $f'_1, \ldots , f'_ r$ map to a regular sequence in $k[x_1, \ldots , x_ n]_{\mathfrak p'}$ generating $I'_{\mathfrak p'}$ by Lemma 15.30.7. Applying Algebra, Lemma 10.135.4 we conclude $S'_{gg'}$ for some $g' \in S$, $g' \not\in \mathfrak q'$ is a global complete intersection over $k$ as desired. $\square$

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