Noetherian Ring with Infinite Krull Dimension

Noetherian Ring with Infinite Krull Dimension

This one is due to Nagata. Consider a polynomial ring    R in countably infinitely many indeterminates over a field  k. Let, for each  n , \mathfrak{p}_n denote the prime ideal  (x_{2^{n-1}},x_{2^{n-1} + 1},...,x_{2^{n}-1}). Then, define the (multiplicative) subset  S = R - \cup_i \mathfrak{p}_i. The localization  A = S^{-1}R will be our counterexample.

We claim that  A is Noetherian. To see this, consider a prime ideal  I \leq A . First, each  \mathfrak{p}_n is maximal and each  f \neq 0 is contained in only finitely many  \mathfrak{p}_n . So,  I is contained in only finitely many  \mathfrak{p}_n . For each  n  I_{\mathfrak{p}_n} is finitely generated. Thus, we can obtain a finite list  a_1,...,a_N of elements of  I that generate each  I_{\mathfrak{p}_n} \subset A_{\mathfrak{p}_n}. Note that, by Hilbert’s basis theorem, A_{\mathfrak{p}_n} is Noetherian. Let  J = (a_1,...,a_N). Observe that  I_\mathfrak{m} = J_\mathfrak{m} for all but finitely many maximal ideals  \mathfrak{m}_1,...,\mathfrak{m}_M, but the  \mathfrak{m}_j are not any of the  \mathfrak{p}_i in which  I is contained. Moreover, for all  j, \exists b_i \in I - \mathfrak{m}_j . Therefore,  (a_1,...,a_N,b_1,...,b_M)   gives  I in each localization and so  I is finitely generated. As all prime ideals of  A are finitely generated, it is Noetherian.

Lastly, for each  n there is a chain (x_{2^{n-1}}),...,(x_{2^{n-1}},x_{2^{n-1} + 1},...,x_{2^{n}-1}) of prime ideals. The supremum then of the lengths of chains of prime ideals in  A is infinite e.g.  A has infinite Krull dimension.


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