I’m trying to think of an example for an integral domain of dimension two with exactly four prime ideals – I fail to find one. Does such a ring exist?
Furthermore, is there a local ring with this property? Is any such ring local?
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Eric Wofsey
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k.stm
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Such a ring might be the coordinate ring of an affine variety of degree 4 over $\mathbb{F}{2}$. The four primes/maximal ideals would be the points of the variety, while dimension 2 means the points live in a dimension/degree 2 extension of $\mathbb{F}{2}$. But, in saying that, I can't immediately think of such a variety/ring. – User0112358 May 24 '17 at 23:25
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@User0112358 Such a ring can't be noetherian, otherwise would have infinitely many primes. – user26857 May 25 '17 at 07:11
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@user26857 do you mean to say a Noetherian ring (integral domain?) necessarily has infinitely many primes? – User0112358 May 25 '17 at 10:53
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@User0112358 I mean that a noetherian integral domain of dimension two has infinitely many primes. – user26857 May 25 '17 at 21:55
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@user26857 okay, you're right. But now I am interested in this :P Is this fact obvious? – User0112358 May 26 '17 at 00:49
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1@User0112358 Obvious not, but pretty standard; see this answer. – user26857 May 26 '17 at 07:14
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@user26857 thanks :) – User0112358 May 26 '17 at 07:38
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I don't know an example off the top of my head, but there is a quite general existence theorem. Namely, if $P$ is any finite poset, then there exists a commutative ring $R$ whose poset of prime ideals is isomorphic to $P$. So in your case, you could take $P$ to be $\{a,b,c,d\}$ with $a<b<c$ and $a<d$ (for instance). You then get a two dimensional ring $R$ with exactly four primes and one minimal prime, and modding out the minimal prime gives a domain. This example is not local, since it has two different maximal ideals ($c$ and $d$), but you could make it local by changing the poset so that $d<c$.
This existence theorem is a special case of an even more general theorem of Hochster characterizing the topological spaces that can be Spec of a ring. See Ring with spectrum homeomorphic to a given topological space.
Eric Wofsey
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1Actually, I was wondering about exactly this question. This is extremely helpful. – k.stm May 25 '17 at 08:50