Lemma 110.40.1. There exists a morphism of affine schemes of finite presentation X \to S and an \mathcal{O}_ X-module \mathcal{F} of finite presentation such that \mathcal{F} is pure relative to S, but not universally pure relative to S.
110.40 Pure not universally pure
Let k be a field. Let
In other words, a power series f \in k[[x, y]] is in R if and only if f(0, y) is a constant. In particular R[1/x] = k[[x, y]][1/x] and R/xR is a local ring with a maximal ideal whose square is zero. Denote R[y] \subset k[[x, y]] the set of power series f \in k[[x, y]] such that f(0, y) is a polynomial in y. Then R \to R[y] is a finite type but not finitely presented ring map which induces an isomorphism after inverting x. Also there is a surjection R[y]/xR[y] \to k[y] whose kernel has square zero. Consider the finitely presented ring map R \to S = R[t]/(xt - xy). Again R[1/x] \to S[1/x] is an isomorphism and in this case S/xS \cong (R/xR)[t]/(xy) maps onto k[t] with nilpotent kernel. There is a surjection S \to R[y], t \longmapsto y which induces an isomorphism on inverting x and a surjection with nilpotent kernel modulo x. Hence the kernel of S \to R[y] is locally nilpotent. In particular S \to R[y] induces a universal homeomorphism on spectra.
First we claim that S is an S-module which is relatively pure over R. Since on inverting x we obtain an isomorphism we only need to check this at the maximal ideal \mathfrak m \subset R. Since R is complete with respect to its maximal ideal it is henselian hence we need only check that every prime \mathfrak p \subset R, \mathfrak p \not= \mathfrak m, the unique prime \mathfrak q of S lying over \mathfrak p satisfies \mathfrak mS + \mathfrak q \not= S. Since \mathfrak p \not= \mathfrak m it corresponds to a unique prime ideal of k[[x, y]][1/x]. Hence either \mathfrak p = (0) or \mathfrak p = (f) for some irreducible element f \in k[[x, y]] which is not associated to x (here we use that k[[x, y]] is a UFD – insert future reference here). In the first case \mathfrak q = (0) and the result is clear. In the second case we may multiply f by a unit so that f \in R[y] (Weierstrass preparation; details omitted). Then it is easy to see that R[y]/fR[y] \cong k[[x, y]]/(f) hence f defines a prime ideal of R[y] and \mathfrak mR[y] + fR[y] \not= R[y]. Since S \to R[y] induces a universal homeomorphism on spectra we deduce the desired result for S also.
Second we claim that S is not universally relatively pure over R. Namely, to see this it suffices to find a valuation ring \mathcal{O} and a local ring map R \to \mathcal{O} such that \mathop{\mathrm{Spec}}(R[y] \otimes _ R \mathcal{O}) \to \mathop{\mathrm{Spec}}(\mathcal{O}) does not hit the closed point of \mathop{\mathrm{Spec}}(\mathcal{O}). Equivalently, we have to find \varphi : R \to \mathcal{O} such that \varphi (x) \not= 0 and v(\varphi (x)) > v(\varphi (xy)) where v is the valuation of \mathcal{O}. (Because this means that the valuation of y is negative.) To do this consider the ring map
defined in the obvious way. We can find a valuation ring \mathcal{O} dominating the localization of the right hand side at the maximal ideal (y^{-1}, x) and we win.
Proof. See discussion above. \square
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