Lemma 37.24.7. Let f : X \to Y be a morphism of schemes. Assume that Y is irreducible and f is of finite type. There exists a diagram
\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_ V \ar[r] \ar[d] & X \ar[d]^ f \\ Y' \ar[r]^ g & V \ar[r] & Y }
where
V is a nonempty open of Y,
X_ V = V \times _ Y X,
g : Y' \to V is a finite universal homeomorphism,
X' = (Y' \times _ Y X)_{red} = (Y' \times _ V X_ V)_{red},
g' is a finite universal homeomorphism,
Y' is an integral affine scheme,
f' is flat and of finite presentation, and
the generic fibre of f' is geometrically reduced.
Proof.
Let V = \mathop{\mathrm{Spec}}(A) be a nonempty affine open of Y. By assumption the Jacobson radical of A is a prime ideal \mathfrak p. Let K = \kappa (\mathfrak p). Let p be the characteristic of K if positive and 1 if the characteristic is zero. By Varieties, Lemma 33.6.11 there exists a finite purely inseparable field extension K'/K such that (X_{K'})_{red} is geometrically reduced over K'. Choose elements x_1, \ldots , x_ n \in K' which generate K' over K and such that some p-power of x_ i is in A/\mathfrak p. Let A' \subset K' be the finite A-subalgebra of K' generated by x_1, \ldots , x_ n. Note that A' is a domain with fraction field K'. By Algebra, Lemma 10.46.7 we see that A \to A' induces a universal homeomorphism on spectra. Set Y' = \mathop{\mathrm{Spec}}(A'). Set X' = (Y' \times _ Y X)_{red}. The generic fibre of X' \to Y' is (X_{K'})_{red} by Lemma 37.24.6 which is geometrically reduced by construction. Note that X' \to X_ V is a finite universal homeomorphism as the composition of the reduction morphism X' \to Y' \times _ Y X (see Morphisms, Lemma 29.45.6) and the base change of g. At this point all of the properties of the lemma hold except for possibly (7). This can be achieved by shrinking Y' and hence V, see Morphisms, Proposition 29.27.1.
\square
Comments (2)
Comment #8432 by Ben Church on
Comment #9056 by Stacks project on