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

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    $\begingroup$ Probably should be asked on MSE. $\endgroup$
    – Bernie
    Commented Dec 9, 2018 at 4:03
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    $\begingroup$ 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. $\endgroup$
    – Asvin
    Commented Dec 10, 2018 at 9:00

2 Answers 2


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.

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


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