Is a scheme Noetherian if its topological space and its stalks are?

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

• Probably should be asked on MSE. Dec 9, 2018 at 4:03
• I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before. Dec 10, 2018 at 9:00

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 and 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.

• Can you give some motivation (when you are free) to think of this example.. Dec 9, 2018 at 13:12
• @PraphullaKoushik I edited my answer to include some motivation behind the construction. Dec 9, 2018 at 20:04
• Thanks for the positive response.. This makes so much sense... Thanks.. Dec 9, 2018 at 20:32

A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.