Lemma 109.39.1. 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 $S$.

## 109.39 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 $S \to R[y]$ is a universal homeomorphism.

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 $S \to R[y]$ is a universal homeomorphism 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)