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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