Lemma 110.63.1. There exists a reduced scheme $X$ and a schematically dense open $U \subset X$ such that some weakly associated point $x \in X$ is not in $U$.

## 110.63 Weakly associated points and scheme theoretic density

Let $k$ be a field. Let $R = k[z, x_ i, y_ i]/(z^2, zx_ iy_ i)$ where $i$ runs over the elements of $\mathbf{N}$. Note that $R = R_0 \oplus M_0$ where $R_0 = k[x_ i, y_ i]$ is a subring and $M_0$ is an ideal of square zero with $M_0 \cong R_0/(x_ iy_ i)$ as $R_0$-module. The prime $\mathfrak p = (z, x_ i)$ is weakly associated to $R$ as an $R$-module (Algebra, Definition 10.66.1). Indeed, the element $z$ in $R_\mathfrak p$ is nonzero but annihilated by $\mathfrak pR_\mathfrak p$. On the other hand, consider the open subscheme

We claim that $U \subset S$ is scheme theoretically dense (Morphisms, Definition 29.7.1). To prove this it suffices to show that $\mathcal{O}_ S \to j_*\mathcal{O}_ U$ is injective where $j : U \to S$ is the inclusion morphism, see Morphisms, Lemma 29.7.5. Translated back into algebra, we have to show that for all $g \in R$ the map

is injective. Write $g = g_0 + m_0$ with $g_0 \in R_0$ and $m_0 \in M_0$. Then $R_ g = R_{g_0}$ (details omitted). Hence we may assume $g \in R_0$. We may also assume $g$ is not zero. Now $R_ g = (R_0)_ g \oplus (M_0)_ g$. Since $R_0$ is a domain, the map $(R_0)_ g \to \prod (R_0)_{x_ ig}$ is injective. If $g \in (x_ iy_ i)$ then $(M_0)_ g = 0$ and there is nothing to prove. If $g \not\in (x_ iy_ i)$ then, since $(x_ iy_ i)$ is a radical ideal of $R_0$, we have to show that $M_0 \to \prod (M_0)_{x_ ig}$ is injective. The kernel of $R_0 \to M_0 \to (M_0)_{x_ n}$ is $(x_ iy_ i, y_ n)$. Since $(x_ iy_ i, y_ n)$ is a radical ideal, if $g \not\in (x_ iy_ i, y_ n)$ then the kernel of $R_0 \to M_0 \to (M_0)_{x_ ng}$ is $(x_ iy_ i, y_ n)$. As $g \not\in (x_ iy_ i, y_ n)$ for all $n \gg 0$ we conclude that the kernel is contained in $\bigcap _{n \gg 0} (x_ iy_ i, y_ n) = (x_ iy_ i)$ as desired.

Second example due to Ofer Gabber. Let $k$ be a field and let $R$, resp. $R'$ be the ring of functions $\mathbf{N} \to k$, resp. the ring of eventually constant functions $\mathbf{N} \to k$. Then $\mathop{\mathrm{Spec}}(R)$, resp. $\mathop{\mathrm{Spec}}(R')$ is the Stone-Čech compactification^{1} $\beta \mathbf{N}$, resp. the one point compactification^{2} $\mathbf{N}^* = \mathbf{N} \cup \{ \infty \} $. All points are weakly associated since all primes are minimal in the rings $R$ and $R'$.

**Proof.**
In the first example we have $\mathfrak p \not\in U$ by construction. In Gabber's examples the schemes $\mathop{\mathrm{Spec}}(R)$ or $\mathop{\mathrm{Spec}}(R')$ are reduced.
$\square$

## 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)