Lemma 110.62.1. 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 $U$.
110.62 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
\[ U = \bigcup D(x_ i) \subset \mathop{\mathrm{Spec}}(R) = S \]
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
\[ R_ g \longrightarrow \prod R_{x_ ig} \]
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 compactification1 $\beta \mathbf{N}$, resp. the one point compactification2 $\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. $\square$
[1] Every element $f \in R$ is of the form $ue$ where $u$ is a unit and $e$ is an idempotent. Then Algebra, Lemma 10.26.5 shows $\mathop{\mathrm{Spec}}(R)$ is Hausdorff. On the other hand, $\mathbf{N}$ with the discrete topology can be viewed as a dense open subset. Given a set map $\mathbf{N} \to X$ to a Hausdorff, quasi-compact topological space $X$, we obtain a ring map $\mathcal{C}^0(X; k) \to R$ where $\mathcal{C}^0(X; k)$ is the $k$-algebra of locally constant maps $X \to k$. This gives $\mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(\mathcal{C}^0(X; k)) = X$ proving the universal property.
[2] Here one argues that there is really only one extra maximal ideal in $R'$.
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