Lemma 23.8.8. Let $S$ be a finite type algebra over a field $k$.
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
$S$ 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #1298 by Kestutis Cesnavicius on
Comment #1308 by Johan on
There are also: