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Noetherian ring of infinite Krull dimension

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Noetherian ring of infinite Krull dimension

Hi,

In Eisenbud’s Commutative Algebra with a view towards algebraicPlanetmathPlanetmath topology there is an example of a such ring. I have a question concerning one of the steps to find one (sectionPlanetmathPlanetmath 9.2, exercise 9.6, page 229). I’ll do my best to explain myself if you don’t have the book:

Let kkk be an algebraic closed field and R=k[x1,x2,,xr,]Rksubscriptx1subscriptx2normal-…subscriptxrnormal-…R=k[x_{1},x_{2},...,x_{r},...] be a polynomial ring in infinitely many variables over kkk, and let P1=(x1,,xd(1))subscriptP1subscriptx1normal-…subscriptxd1P_{1}=(x_{1},...,x_{{d(1)}}), P2=(xd(1)+1,,xd(2))subscriptP2subscriptxd11normal-…subscriptxd2P_{2}=(x_{{d(1)+1}},...,x_{{d(2)}}), …, Pm=(xd(m-1)+1,,xd(m)),subscriptPmsubscriptxdm11normal-…subscriptxdmnormal-…P_{m}=(x_{{d(m-1)+1}},...,x_{{d(m)}}),... be an infinite collection of prime idealsPlanetmathPlanetmathPlanetmath made from disjoint subsets of the variables.

Let U=R-m=1PmURsuperscriptsubscriptm1subscriptPmU=R-\bigcup_{{m=1}}^{{\infty}}P_{m}. As we know, this is multiplicative set in RRR and thus we can form the localization of UUU at RRR, denoted by S=R[U-1]SRsuperscriptU1S=R[U^{{-1}}].

The idea is to prove that SSS has infinite Krull dimension. To achieve this, the book uses an exercise: If IRIRI\subset R is an ideal and Im=1PmIsuperscriptsubscriptm1subscriptPmI\subset\bigcup_{{m=1}}^{{\infty}}P_{m}, then IPnIsubscriptPnI\subset P_{n} for some nnn. The problem is that this seems false. Here an example:

I=(xd(1)-1,xd(1)+1)P1P2Isubscriptxd11subscriptxd11subscriptP1subscriptP2I=(x_{{d(1)-1}},x_{{d(1)+1}})\subset P_{1}\cup P_{2} but III is not contained in any of the PisubscriptPiP_{i}. Is there something I am missing?

Thanks in advance.


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