Lemma 23.8.8. Let $S$ be a finite type algebra over a field $k$.

1. for a prime $\mathfrak q \subset S$ the local ring $S_\mathfrak q$ is a complete intersection in the sense of Algebra, Definition 10.135.5 if and only if $S_\mathfrak q$ is a complete intersection in the sense of Definition 23.8.5, and

2. $S$ is a local complete intersection in the sense of Algebra, Definition 10.135.1 if and only if $S$ is a local complete intersection in the sense of Definition 23.8.5.

Proof. Proof of (1). Let $k[x_1, \ldots , x_ n] \to S$ be a surjection. Let $\mathfrak p \subset k[x_1, \ldots , x_ n]$ be the prime ideal corresponding to $\mathfrak q$. Let $I \subset k[x_1, \ldots , x_ n]$ be the kernel of our surjection. Note that $k[x_1, \ldots , x_ n]_\mathfrak p \to S_\mathfrak q$ is surjective with kernel $I_\mathfrak p$. Observe that $k[x_1, \ldots , x_ n]$ is a regular ring by Algebra, Proposition 10.114.2. Hence the equivalence of the two notions in (1) follows by combining Lemma 23.8.3 with Algebra, Lemma 10.135.7.

Having proved (1) the equivalence in (2) follows from the definition and Algebra, Lemma 10.135.9. $\square$

Comment #1298 by Kestutis Cesnavicius on

'acomplete' ---> 'a complete'

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