Lemma 10.135.2. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $g \in S$.

1. If $S$ is a global complete intersection so is $S_ g$.

2. If $S$ is a local complete intersection so is $S_ g$.

Proof. The second statement follows immediately from the first. Proof of the first statement. If $S_ g$ is the zero ring, then it is true. Assume $S_ g$ is nonzero. Write $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $n - c = \dim (S)$ as in Definition 10.135.1. By the remarks following the definition $\dim (S_ g) = n - c$. Let $g' \in k[x_1, \ldots , x_ n]$ be an element whose residue class corresponds to $g$. Then $S_ g = k[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, x_{n + 1}g' - 1)$ as desired. $\square$

Comment #2913 by Dario Weißmann on

"By the remarks above..." is a bit imprecise.

Firstly the remarks cite section 10.59, lemma 10.59.12 would be more accurate. That the local rings at maximal ideals of polynomial algebras over a field have dimension $n$ is Lemma 10.113.1.

Secondly if $g$ is contained in all maximal ideal it is nilpotent (thus $S_g$ zero), follows from Hilbert's Nullstellensatz or more precisely Lemma 10.34.2.

Comment #2942 by on

Sometimes when there is a definition, there are some comments directly following the definition explaining how to think about the definition. We often assume the reader has understood these comments and we'll use them in future use of the concept. Anyway, I've made the comments more clear and I have then referred to them in the proof of this lemma. For changes see here.

Comment #8195 by Ryo Suzuki on

I don't understand why equidimensionality of $S$ directly implies ${\rm dim}(S_g)={\rm dim}(S)$, although I could understand the result is true thanks to Comment #2913.

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