Let $A$ be a ring. If $\operatorname{Spec} A$ is a finite set, then $A$ is a noetherian ring?

Question

$A$ : commutative with unit

$\operatorname{Spec} A$ is finite $\Rightarrow$ $A$ has finitely many maximal ideals $ = \{M_1, M_2, ..., M_n\} $

If $ I_1 \subseteq I_2 \subseteq...\subseteq I_n \subseteq...$ with every $I_j \subseteq M$, $M$ is a maximal ideal of $A$.

Can this imply this ascending chain is stationary?

There are non-noetherian local rings, so you would need something more from the assumptions than just finitely many maximal ideals. — Brian Moehring, May 17 '20 at 05:55
Consider the ring $\mathbb{Z}[x_1,x_2,\ldots]_{(x_1,x_2,\ldots)}$. This ring has one maximal ideal but is not Noetherian. Thus, it is not enough to use that there are finitely many maximal ideals. — Anonymous, May 17 '20 at 05:56
Spec$,A$ generally denotes the set of prime ideals of $A$, — Angina Seng, May 17 '20 at 06:07
Please use \operatorname{Spec} to typeset $\operatorname{Spec}$: this produces the best spacing. — KReiser, May 17 '20 at 06:18

KReiser · Accepted Answer · 2020-05-17T06:33:20.057

6

No. Consider $A=k[x_1,x_2,\cdots]/(x_1,x_2,\cdots)^2$. This has a unique prime ideal $(x_1,x_2,\cdots)$, so $\operatorname{Spec} A$ is a single point. On the other hand, $A$ is not Noetherian since this ideal is not finitely generated.

edited May 17 '20 at 06:33

answered May 17 '20 at 06:17

KReiser

65,137

In this case, is Spec $A$ a finite set ? – NEMO May 17 '20 at 06:23
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$\operatorname{Spec} A$ is one point. I've edited for clarity. – KReiser May 17 '20 at 06:30
How does one see $(x_1,x_2,\dots)$ is the only prime ideal? It seems like $A$ is effectively all linear polynomials in infinitely many variables, with the relation $x_ix_j=0$ for any $i,j$. Why isn't something like $(x_1+x_2)$ also prime in $A$? – Adelaide Dokras May 17 '20 at 06:36
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@AdelaideDokras The nilradical of a ring is contained in every prime ideal. Every polynomial with no constant term in $A$ is nilpotent (it squares to zero), so all prime ideals must contain $(x_1,x_2,\cdots)$. On the other hand, $A/(x_1,x_2,\cdots)\cong k$, so by the correspondence theorem there are no proper ideals containing $(x_1,x_2,\cdots)$. So this is the unique prime ideal. – KReiser May 17 '20 at 06:45
I see now, thanks. – Adelaide Dokras May 17 '20 at 06:57
This is great. Thank you very much. – NEMO May 17 '20 at 07:07
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@NEMO After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – KReiser May 17 '20 at 07:18

Let $A$ be a ring. If $\operatorname{Spec} A$ is a finite set, then $A$ is a noetherian ring?

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