Exercise about Krull Dimension/Eisenbud, Exercise 10.1

Question

I can't solve this exercise. If someone can help me, thanks a lot.

Let $R$ be a Noetherian ring, and $x$ an indeterminate. Prove that $\dim R[x,x^{-1}]=\dim R+1$.

Thank for your answers!

Answer 1

One knows that $\dim R[x]=\dim R+1$ (Eisenbud, Corollary 10.13(b)).

The prime ideals of $R[x,x^{-1}]$ correspond to the prime ideals of $R[x]$ which don't contain $x$, so $\dim R[x,x^{-1}]\le \dim R[x]=\dim R+1$.

On the other side, a chain of prime ideals in $R$ extends to a chain of prime ideals in $R[x,x^{-1}]$ if one associates to a prime $p$ its extension $p[x,x^{-1}]$; note that even if $p$ is maximal, $p[x,x^{-1}]$ is not maximal since $R[x,x^{-1}]/p[x,x^{-1}]\simeq (R/p)[x,x^{-1}]$, so $\dim R[x,x^{-1}]\ge\dim R+1$.