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
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.
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