# A scheme that can be covered by finitely many affine open subschemes with noetherian topology is not necessarily a noetherian scheme?

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.

• @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/… Apr 26, 2018 at 21:11
• @Ti Wen: The link you are referring to answers your question. In particular, it gives you such a scheme which is additionally affine. Apr 26, 2018 at 21:14
• @PavelČoupek Thanks! How do you show that scheme is non-noetherian? Apr 26, 2018 at 21:16
• @TiWen sorry I misread this Apr 26, 2018 at 21:17
• @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. Apr 26, 2018 at 21:50

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]