Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?



Is a scheme Noetherian if its topological space and its stalks are?
This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at *every* closed [rational] point":


**Example.** Let $k$ be an infinite field, let $R = k[x]$, and for each $\alpha \in k$ let $R\_\alpha = R[y\_\alpha]/((x-\alpha)y\_\alpha,y\_\alpha^2)$. Then $R\_\alpha$ is an affine line with an embedded prime $\mathfrak p\_\alpha = (x-\alpha,y\_\alpha)$ at $x = \alpha$, sticking out in the $y\_\alpha$-direction. Finally, let
$$R\_\infty = \bigotimes\_{\alpha \in k} R\_\alpha = \operatorname\*{colim}\_{\substack{\longrightarrow\\I \subseteq k\\\text{finite}}} \bigotimes\_{\alpha \in I} R\_\alpha$$
be their tensor product over $R$ (*not* over $k$); that is
$$R\_\infty = \frac{k[x]\left[\{y\_\alpha\}\_{\alpha \in k}\right]}{\sum\_{\alpha \in k}((x-\alpha)y\_\alpha, y\_\alpha^2)}.$$
This is not a Noetherian ring, because the radical $\mathfrak r = (\{y\_\alpha\}\_{\alpha \in k})$ is not finitely generated. But $\operatorname{Spec} R\_\infty$ agrees as a topological space with $\operatorname{Spec} R\_\infty^{\operatorname{red}} = \mathbb A^1\_k$, hence $|\!\operatorname{Spec} R\_\infty|$ is a Noetherian topological space.


On the other hand, the map $R \to R\_\alpha$ is an isomorphism away from $\alpha$, and similarly $R\_\alpha \to R\_\infty$ induces isomorphisms on the stalks at $\alpha$. Thus, the stalk $(R\_\infty)\_{\mathfrak q\_\alpha} = (R\_\alpha)\_{\mathfrak p\_\alpha}$ at $\mathfrak q\_\alpha = \mathfrak p\_\alpha R\_\infty + \mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $\mathfrak r$ is just $R\_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R\_\infty$ are Noetherian. $\square$


**Remark.** As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I\_0 \subseteq I\_1 \ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $\mathcal O\_{X,x}$ for the generic points $x$ of $V$.


However this is not quite true, because of embedded points. The precise statement [Tags [0311](https://stacks.math.columbia.edu/tag/0311) and [02M3](https://stacks.math.columbia.edu/tag/02M3)] is that the inclusion $I\_{i-1} \subseteq I\_i$ is an equality if and only if the same holds for the localisation at every associated point of $I\_i$. Now you run into trouble if the $I\_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $\mathcal O\_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.


