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Here is the question:

Let $A$ be a commutative ring with unit, $X=\mathrm{Spec}A$, $U_i$s be quasi-compact open sets of $X$ such that $\emptyset=\cap_{i\in I}U_i$, then there is a finite subset $I_0$ of $I$ such that $\emptyset=\cap_{i\in I_0}U_i$.

I just did a more simpler case, that is when all $U_i$s are distinguished open set $D(f_i)$. Following is a proof: Let $S$ be the multicative set of finite product of $f_i$s, then $A_S=0$. Thus we have a finite product $f_1f_2\cdots f_n=0$ (here we allow $f_i=f_j$), $\cap_{i=1}^nD(f_i)=\emptyset$.

I do not know how to do the previous case. Could anyone give some advices?

Thanks.

Edit: Dear Pierre-Yves Gaillard, here is the story I found the problem:)

A topological space is said spectral space if it is homeomorphic to the spectrum of a commutative ring. And proposition 8 in this paper(M. Hochster (1969). "Prime ideal structure in commutative rings." Trans. Amer. Math. Soc., 142 43—60) implies this statement is true.

I copy the content of proposition 8 as follow:

Proposition 8. Let $X$ be spectral. Retopologize $X$ by taking as a basis for the closed sets the quasi-compact open sets of X. Then $X$ with this new topology is spectral, and the new order induced on $X$ by this topology is precisely the reverse of the original order.

I found this problem from a talk with my friend, and I wonder if there is a simple proof about this special case(i.e. I just assume all $U_i$s are q.c. not the general closed sets of the inverse topology) before I read the paper. When I try to prove it, I found it is not easy for me, and I just solve the case of distinguished opens.

(aside:I know there must be many wrong about the syntax/grammar, I will appreciate that if you improve this post.)

Update: Using Alexander subbase theorem, we can check the quasi-compactness at the level of a subbase! So in fact, we are done! Today I know this powerful theorem.

wxu
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1 Answers1

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I haven't solved it; this is just a suggestion. I'm not strong on algebra, so I'd first try to translate back into ideals:

"Let $(I_i)_{i\in S}$ be a collection of finitely generated ideals. If every prime ideal contains $I_i$ for some $i$ then there exists a finite subset $S_0\subset S$ such that every prime ideal contains $I_i$ for some $i\in S_0$."

(Though you might need a stronger consequence of quasi-compactness.)

Now we can try to generalize the case you've done already. If some product $I_{i_1} \cdots I_{i_k}$ is nilpotent we're done: every prime ideal contains every nilpotent ideal, and hence contains some $I_{i_j}$. Pick a maximal ideal not containing any ideal of the form $\prod_i I_i^{t_i}$ where $t_i$ are non-negative integers with finitely many non-zero terms. We would like to show that this is prime, for example by showing that the set of ideals containing some $\prod_i I_i^{t_i}$ is an Oka family (see this answer or ch 4 section 1.3 of the CRing textbook).

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Colin McQuillan
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  • Zorn's lemma can apply, it has maximal elements. It remains to show that maximal elements are prime. – wxu Sep 21 '11 at 15:31
  • How about this? Suppose $M$ contains $AB$ and there exist $a\in A\setminus M$ and $b\in B\setminus M$. Then $(M,a)$ contains some $\prod I_i^{A_i}$ and $(M,b)$ contains some $\prod I_i^{B_i}$. But $M$ contains $(M,a)(M,b)$, which contains $\prod I_i{A_i+B_i}$. So maybe quasicompactness isn't needed? – Colin McQuillan Sep 21 '11 at 16:57
  • Right! Quasicompactness uses to show there exists maximal elements. – wxu Sep 22 '11 at 00:39
  • That $I_i$ is finitely generated is equivalent to $U_i$ is quasicompact, and we use the finiteness of $\prod I_i$ to show there exists maximal elements. Finally, to show $M$ is prime as you said, if $x,y\notin M$, but $xy\in M$, then $M\supset (M,x)(M,y)\supset \prod I_i$, contradiction. Thus $M$ is a prime ideal. +1. – wxu Sep 22 '11 at 00:58