Let $A= \mathbb{Z}[x]$ and $ m=(2,x)$. Find the Krull dimension of $A_m$.

Question

Let $A=\mathbb{Z}[x]$ and $m=(2,x)$.
$1$. Then what is the Krull dimension of $A_m$?
$2$. If $B=A_m/(x^2-125)$ what is the Krull dimension of $ B $?

Any suggestions?

Answer 1

Note that $\dim A_m = \text{ht}(m) \le 2$ by Krull's Altitude Theorem. Can you find a chain of prime ideals (in $A$) of length $2$ contained in $m$? It suffices to find a nonzero prime ideal strictly contained in $m$.

(Alternatively, since $\mathbb{Z}[x]$ is a UFD, if $\text{ht}(m) = 1$, then $m$ would be principal - show this is not true).

For the second part, note that $x^2 - 125 \equiv 1 \pmod{m}$, so $x^2 - 125$ is not in the maximal ideal of $A_m$, but $A_m$ is a local ring...