Background: A Noetherian ring is said to be regular if its localizations at all prime (or maximal) ideals are regular local rings. Without this assumption, there are counter-examples.
Thanks.
Background: A Noetherian ring is said to be regular if its localizations at all prime (or maximal) ideals are regular local rings. Without this assumption, there are counter-examples.
Thanks.
No. An example is given in K. Fujita, Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181-185.
Here's my favorite example (is this the one nosr refers to in his comment? I'm pretty sure it's also due to Nagata). Let $k$ be a field and $A=k[x_1, x_2, x_3, \ldots]$ be polynomial ring in countably many variables. Let $P_1 := (x_1)$, $P_2 := (x_2, x_3)$, $P_3 := (x_4, x_5, x_6)$, and in general $P_n$ is generated by the "next" $n$ variables. That is, $P_n := \left(x_{{n \choose 2} + 1}, x_{{n \choose 2} + 2}, \ldots, x_{{n+1} \choose 2}\right)$. Let $W := A \setminus \bigcup_{n=1}^\infty P_n$, and let $R := W^{-1}A$. Then every prime ideal of $R$ is in some $P_nR$, each of which is a maximal ideal of $R$, and $R_{P_n R} \cong k[y_1, \dotsc, y_n]_{(y_1, \dotsc, y_n)}$ is certainly a regular local ring. Hence $R$ is a regular Noetherian ring. But as it has essentially polynomial rings of every dimension as localizations, $R$ has infinite dimension.
On the other hand, every Noetherian local ring has finite Krull dimension.