Locally ringed space with noetherian stalks and a non-coherent structural sheaf

Question

I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?

Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).

Answer 1

Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.

Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=\text{Spec}\ k[t]_{\langle t\rangle}$, and let $Z$ be the singleton set $\{z\}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.

Consider the sheaf of $\mathcal{O}_{Y}$-modules $i_*i^{-1}\mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $y\in U$ is the zero module. Also, for every $z\in Z$, the stalk at $z$ equals $\mathcal{O}_{Y,z}$.

Now form the sheaf of $\mathcal{O}_Y$-modules $\mathcal{O}_X:=\mathcal{O}_Y \oplus \left( i_*i^{-1}\mathcal{O}_Y\cdot \epsilon \right)$, where $\epsilon$ is just a placeholder. Give this the unique structure of $\mathcal{O}_Y$-algebra such that $\epsilon\cdot \epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $\mathcal{O}_Y$-algebras, $$\mathcal{O}_Y \hookrightarrow \mathcal{O}_Y \oplus \left( i_*i^{-1}\mathcal{O}_Y\cdot \epsilon \right), \ \ f\mapsto (f,0\cdot \epsilon).$$

For every $y\in Y\setminus Z$, the stalk $\mathcal{O}_{X,y}$ equals $\mathcal{O}_{Y,y}$ as an $\mathcal{O}_{Y,y}$-algebra. Also, for every $z\in Z$, the $\mathcal{O}_{Y,z}$-algebra $\mathcal{O}_{X,z}$ equals $\mathcal{O}_{Y,z}\oplus \mathcal{O}_{Y,z}\cdot \epsilon$ with $\epsilon\cdot \epsilon = 0$. In every case, the stalk is a Noetherian local ring.

In the special case that $Y$ equals $\text{Spec}\ k[t]_{\langle t\rangle}$ and $Z$ equals the singleton of the closed point, then also $\mathcal{O}_X(Y)$ equals $k[t]_{\langle t \rangle} \oplus k[t]_{\langle t \rangle}\cdot \epsilon$, which is Noetherian. Also $\mathcal{O}_X(Y\setminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.

There is a natural retraction of the above algebra homomorphism, $$\mathcal{O}_X\to \mathcal{O}_Y, \ \ (f,g\cdot \epsilon) \mapsto f.$$ Consider the $\mathcal{O}_X$-module morphism from $\mathcal{O}_X$ to itself that multiplies by the global section $(0,1\cdot \epsilon)$. Denote the kernel sheaf by $\mathcal{K}$. Denote by $\mathcal{K}'$ the associated $\mathcal{O}_Y$-module $\mathcal{K}\otimes_{\mathcal{O}_X}\mathcal{O}_Y$.

If $\mathcal{K}$ is locally finitely generated as an $\mathcal{O}_X$-module, then also $\mathcal{K}'$ is locally finitely generated as an $\mathcal{O}_Y$-module. However, $\mathcal{K}'$ is the extension by zero of the structure sheaf on $Y\setminus Z$. This is not a finitely generated $\mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $\mathcal{K}'$ on $U$ are zero, yet they are nonzero for $U\setminus Z$. Thus, every map $\mathcal{O}_U^{\oplus n}\to \mathcal{K}'|_U$ is the zero map, yet the restriction of $\mathcal{K}'|_U$ to $U\setminus Z$ is nonzero.

Answer 2

A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.

(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)

Answer 3

Start with any ring $A$ which is locally Noetherian (in the sense that $A_\mathfrak{p}$ is Noetherian for every prime $\mathfrak{p}$) but not Noetherian. For example, $A$ could be an infinite product of fields. Let $I$ be an ideal of $A$ which is not finitely generated. Then $B = A[x]/(xI)$ is locally Noetherian. But the annihilator of $x$ in $B$ is $IB$, which is not finitely generated. Thus $B$ is locally Noetherian but not coherent.