# Show a coordinate ring is not Artinian

Show that $$\mathbb{R}[x,y]/(x^2+y^2-1)$$ is not Artinian.

I am told to use my geometric intuition to prove it doesn't satisfy the d.c.c., but my intuition isn't very strong when it comes to coordinate rings. I know this ring is what we get by restricting real-valued polynomial functions to the unit circle and that polynomials equal on the circle are equal in the ring.

But how do I use this to construct a chain of ideals that is strictly decreasing?

$$(x) \supset (x^2) \supset (x^3) \cdots$$

• I was thinking of this chain as well. I just couldn't justify it intuitively, although I don't see why it shouldn't work. – Isomorphic Twin Nov 7 '18 at 16:53
• No I couldn't understand what the geometric intuition is/ should be. I just grabbed the easiest solution that would do the job ... – Richard Martin Nov 7 '18 at 16:55
• @IsomorphicTwin In an integral domain, any nonzero nonunit $x$ will satisfy this property. (It doesn't have to do with it being one of the indeterminates.) But again, that depends on your confidence it is a domain. – rschwieb Nov 7 '18 at 17:29

Let $$p_1,p_2,\cdots$$ be a countably infinite collection of distinct closed points of this variety and $$m_1,m_2,\cdots$$ their corresponding maximal ideals. Then $$m_1\supset m_1m_2 \supset m_1m_2m_3 \supset \cdots$$ and this is an infinite descending chain of ideals that never stabilizes.

The point is that a descending chain of ideals corresponds to an increasing chain of closed subvarieties corresponding to those ideals, by the Galois connection between ideals and closed subvarieties. So looking for a chain of ideals which doesn't satisfy DCC is the same as looking for a chain of subvarieties which doesn't satisfy ACC.

There are many good explanations already, and this answer is similar in spirit to KReiser's, but I think the explanation is different enough that it merits posting. Also Richard Martin's answer is probably the easiest way to see that the ring isn't Artinian, but I thought I'd give this answer which is more geometric.

If $$A$$ is Artinian, then it has finitely many maximal ideals. I.e., $$\newcommand\Spec{\operatorname{Spec}}\Spec A$$ has finitely many closed points. Why is this? Well, let $$\newcommand\mm{\mathfrak{m}}\{\mm_n\}_{n\in\Bbb{N}}$$ be a sequence of distinct maximal ideals. Let $$I_n = \bigcap_{i=0}^n \mm_i$$. $$I_n$$ is a descending chain of ideals in $$A$$ corresponding to taking the sequence of closed sets $$F_n = \{\mm_i : 0\le i \le n\}$$ in $$\Spec A$$. Since all the ideals in the sequence are distinct, all the closed sets $$F_n$$ are distinct as well, so $$I_n$$ is an infinite chain of descending ideals that fails to stabilize. Contradiction.

Now the unit circle in $$\Bbb{R}^2$$ has infinitely many closed points, so its coordinate ring can't be Artinian.

Do you know already that it is a domain?

If so, then an Artinian domain is a field, but this is clearly not a field. You have, for example the nontrivial ideal $$(x)$$.

• I'm afraid not. I'm told to either give a chain that doesn't satisfy the d.c.c. or alternatively show it has infinitely many maximal ideals. – Isomorphic Twin Nov 7 '18 at 16:57