Definition: A noetherian scheme is a scheme that admits a finite covering by open affine subsets $\operatorname {Spec} A_{i}$, where $A_{i}$ are noetherian rings.

I wonder if there is a non-noetherian scheme that satisfies the following slightly weakened conditions

A scheme that admits a finite covering by open affine subsets $\operatorname {Spec} A_{i}$, where $\operatorname {Spec} A_{i}$ are noetherian topological spaces.

  • $\begingroup$ @JonasLenz probably not...i meant $\operatorname {Spec} A_i$ is a noetherian topological space...in this case see the counterexample..math.stackexchange.com/questions/7392/… $\endgroup$
    – No One
    Apr 26, 2018 at 21:11
  • 1
    $\begingroup$ @Ti Wen: The link you are referring to answers your question. In particular, it gives you such a scheme which is additionally affine. $\endgroup$ Apr 26, 2018 at 21:14
  • $\begingroup$ @PavelČoupek Thanks! How do you show that scheme is non-noetherian? $\endgroup$
    – No One
    Apr 26, 2018 at 21:16
  • $\begingroup$ @TiWen sorry I misread this $\endgroup$
    – Jonas Lenz
    Apr 26, 2018 at 21:17
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    $\begingroup$ @TiWen: Short answer: If it was Noetherian, then every affine open subset would have to be given by a Noetherian ring. Long answer: I added that as an answer. $\endgroup$ Apr 26, 2018 at 21:50

3 Answers 3


As you yourself point out in the comments, there are non-Noetherian rings $A$ such that $\mathrm{Spec}\,A$ is a Noetherian topological space. An example from Martin Brandenburg of this is here.

Following up on your question in comments: In order to conclude that $\mathrm{Spec}\,A$ in indeed non-Noetherian scheme, i.e. that there is no finite affine open cover by spectra of Noetherian rings, it's enough to show the following:

Claim: A scheme $X$ is locally Noetherian if and only if for every open affine subset $U=\mathrm{Spec}\,B \subseteq X$, $B$ is a Noetherian ring.

Here "locally Noetherian" means "admits an affine open cover by spectra of Noetherian rings" (i.e. the cover need not be finite). This is clearly enough, because if $\mathrm{Spec}\,A$ is Noetherian, it's clearly locally Noetherian and thus, for the affine subset $\mathrm{Spec}\,A$ we would have to have $A$ a Noetherian ring.

The statement of the Claim can be broken to several mostly technical exercises. The first of these is an affine-local principle that R. Vakil calls affine communication lemma:

Affine communication lemma. Let $\mathbb{P}$ be a property that can be stated about an affine open set of a scheme $X$. Assume further that

  1. Given $U \subseteq X$ affine open set having property $\mathbb{P}$ and a distinguished open set $D \subseteq U,$ $D$ has property $\mathbb{P}$ as well.

  2. Given $U \subseteq X$ affine open set and its open cover $U=\bigcup_i D_i$ by distinguished open subsets, all of $D_i$ having property $\mathbb{P},$ $U$ also have property $\mathbb{P}$.

If the property $\mathbb{P}$ holds for an affine open cover of $X$, then it holds for all affine open subsets of $X$.

After accepting/proving the lemma, it thus remains that the property "$U=\mathrm{Spec}\,B$ for a Noetherian ring $B$" satisfies the assumptions of the lemma. That boils down to two claims of commutative algebra:

  1. Given a Noetherian ring $B$ and $f \in B$, the ring $B_f$ is Noetherian. (This one is actually clear)
  2. Given a ring $B$ and collection of elements $f_1, \dots , f_k \in B$ such that $(f_1, f_2, \dots, f_k)=B$ and such that $B_{f_i}$ is Noetherian for every $i$, the ring $B$ is itself Noetherian.

As Pavel Čoupek said, any non-Noetherian ring whose spectrum is a Noetherian space gives a counterexample. Here's an example where it is very easy to verify this. Let $k$ be a field and let $A=k[x_1,x_2,\dots]/(x_1^2,x_2^2,\dots)$. Then $A$ has only one prime ideal, since every prime idea must contain $x_n$ for all $n$, and the quotient $A/(x_1,x_2,\dots)\cong k$ is already a field. So $\operatorname{Spec} A$ is a Noetherian space. On the other hand, $A$ is not Noetherian, since the unique prime ideal is not finitely generated. Moroever, $\operatorname{Spec} A$ cannot be covered by open subschemes that are spectra of Noetherian rings, since the only nonempty open subset is $\operatorname{Spec} A$ itself.


The following reassuring result simplifies Pavel's and Eric's answers: given a ring $A$ we have

$\operatorname {Spec} A$ is a noetherian scheme $\iff$ $A$ is a noetherian ring

[EGA I, Proposition (6.1.3), page 141]


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