# 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.