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. $\square$

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)'.

There are also:

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