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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?

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

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  • $\begingroup$ 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. $\endgroup$ – Isomorphic Twin Nov 7 '18 at 16:53
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    $\begingroup$ No I couldn't understand what the geometric intuition is/ should be. I just grabbed the easiest solution that would do the job ... $\endgroup$ – Richard Martin Nov 7 '18 at 16:55
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    $\begingroup$ @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. $\endgroup$ – rschwieb Nov 7 '18 at 17:29
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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.

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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.

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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 $(x)$.

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  • $\begingroup$ 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. $\endgroup$ – Isomorphic Twin Nov 7 '18 at 16:57

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