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The Stacks project

Lemma 10.136.11. Let R be a ring. Let S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) be a relative global complete intersection (Definition 10.136.5). There exist a finite type \mathbf{Z}-subalgebra R_0 \subset R such that f_ i \in R_0[x_1, \ldots , x_ n] and such that

S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)

is a relative global complete intersection.

Proof. Let R_0 \subset R be the \mathbf{Z}-algebra of R generated by all the coefficients of the polynomials f_1, \ldots , f_ c. Let S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c). Clearly, S = R \otimes _{R_0} S_0. Pick a prime \mathfrak q \subset S and denote \mathfrak p \subset R, \mathfrak q_0 \subset S_0, and \mathfrak p_0 \subset R_0 the primes it lies over. Because \dim (S \otimes _ R \kappa (\mathfrak p) ) = n - c we also have \dim (S_0 \otimes _{R_0} \kappa (\mathfrak p_0)) = n - c, see Lemma 10.116.5. By Lemma 10.125.6 there exists a g \in S_0, g \not\in \mathfrak q_0 such that all nonempty fibres of R_0 \to (S_0)_ g have dimension \leq n - c. As \mathfrak q was arbitrary and \mathop{\mathrm{Spec}}(S) quasi-compact, we can find finitely many g_1, \ldots , g_ m \in S_0 such that (a) for j = 1, \ldots , m the nonempty fibres of R_0 \to (S_0)_{g_ j} have dimension \leq n - c and (b) the image of \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_0) is contained in D(g_1) \cup \ldots \cup D(g_ m). In other words, the images of g_1, \ldots , g_ m in S = R \otimes _{R_0} S_0 generate the unit ideal. After increasing R_0 we may assume that g_1, \ldots , g_ m generate the unit ideal in S_0. By (a) the nonempty fibres of R_0 \to S_0 all have dimension \leq n - c and we conclude. \square


Comments (2)

Comment #3255 by Dario Weißmann on

There are typos: In the statement 'relative global intersection' and directly above the lemma: '...relative complete intersections' should both be 'relative global complete intersection(s)'.

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  • 2 comment(s) on Section 10.136: Syntomic morphisms

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