Lemma 10.136.10. Let R be a ring. Let S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c). We will find h \in R[x_1, \ldots , x_ n] which maps to g \in S such that
S_ g = R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, hx_{n + 1} - 1)
is a relative global complete intersection with a presentation as in Definition 10.136.5 in each of the following cases:
Let I \subset R be an ideal. If the fibres of \mathop{\mathrm{Spec}}(S/IS) \to \mathop{\mathrm{Spec}}(R/I) have dimension n - c, then we can find (h, g) as above such that g maps to 1 \in S/IS.
Let \mathfrak p \subset R be a prime. If \dim (S \otimes _ R \kappa (\mathfrak p)) = n - c, then we can find (h, g) as above such that g maps to a unit of S \otimes _ R \kappa (\mathfrak p).
Let \mathfrak q \subset S be a prime lying over \mathfrak p \subset R. If \dim _{\mathfrak q}(S/R) = n - c, then we can find (h, g) as above such that g \not\in \mathfrak q.
Proof.
Ad (1). By Lemma 10.125.6 there exists an open subset W \subset \mathop{\mathrm{Spec}}(S) containing V(IS) such that all fibres of W \to \mathop{\mathrm{Spec}}(R) have dimension \leq n - c. Say W = \mathop{\mathrm{Spec}}(S) \setminus V(J). Then V(J) \cap V(IS) = \emptyset hence we can find a g \in J which maps to 1 \in S/IS. Let h \in R[x_1, \ldots , x_ n] be any preimage of g.
Ad (2). By Lemma 10.125.6 there exists an open subset W \subset \mathop{\mathrm{Spec}}(S) containing \mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p)) such that all fibres of W \to \mathop{\mathrm{Spec}}(R) have dimension \leq n - c. Say W = \mathop{\mathrm{Spec}}(S) \setminus V(J). Then V(J \cdot S \otimes _ R \kappa (\mathfrak p)) = \emptyset . Hence we can find a g \in J which maps to a unit in S \otimes _ R \kappa (\mathfrak p) (details omitted). Let h \in R[x_1, \ldots , x_ n] be any preimage of g.
Ad (3). By Lemma 10.125.6 there exists a g \in S, g \not\in \mathfrak q such that all nonempty fibres of R \to S_ g have dimension \leq n - c. Let h \in R[x_1, \ldots , x_ n] be any element that maps to g.
\square
Comments (1)
Comment #717 by Keenan Kidwell on
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