# Noetherian ring of infinite Krull dimension

Hi,

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

Let $k$ be an algebraic closed field and $R=k[x_{1},x_{2},...,x_{r},...]$ be a polynomial ring in infinitely many variables over $k$, and let $P_{1}=(x_{1},...,x_{{d(1)}})$, $P_{2}=(x_{{d(1)+1}},...,x_{{d(2)}})$, …, $P_{m}=(x_{{d(m-1)+1}},...,x_{{d(m)}}),...$ be an infinite collection of prime ideals made from disjoint subsets of the variables.

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

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

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