In Corollary 11.12 of Atiyah-Macdonald, it says that in a Noetherian ring every prime ideal has finite height. It seems to come directly from Proposition 11.10, which says if $A$ is a Noetherian local ring, $\dim A \leq d(A)$, where $d(A)$ is the degree of the characteristic polynomial of $A$, hence finite. It seems like from that proposition you can only conclude that any prime ideal in a Noetherian local ring has finite height. I don't see how you can reach Corollary 11.12.
Every prime ideal has finite height in a Noetherian ring
commutative-algebra
Best Answer
One should be careful not to confuse the height of an ideal with the Krull dimension of the ring. A famous example of Nagata shows that a Noetherian ring can indeed have infinite Krull dimension, precisely because it has prime ideals which have arbitrarily large height. See this question for details of Nagata's example.
However, the height of any prime ideal $\mathfrak{p}$ in a Noetherian ring $R$ can still never be infinite. To see this, as the comments mention, one can combine Proposition 11.10 in Atiyah-Macdonald with the fact that $\operatorname{ht}\mathfrak{p}=\dim R_{\mathfrak{p}}$, which follows from their Corollary 3.13, to conclude that $\operatorname{ht}\mathfrak{p}$ is finite.