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 counterexamples.
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 counterexamples. Thanks. 


No. An example is given in K. Fujita, Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181185. 


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. 

