# Krull dimension of $R[X]/(f(X))$ for $f(X)$ monic

How can I prove that the Krull dimension of $R[X]/(f(X))$, for $R$ a finitely generated noetherian integral domain and $f(X)$ monic, is equal to the Krull dimension of $R$?

I don't even know where to start, since even to use Noether normalization I would need $(f(X))$ to be prime, right? Any help would be great.

• Do you mean Krull dimension? – arsmath Jan 29 '15 at 17:27
• @arsmath yes. edited. – user198182 Jan 29 '15 at 17:28
• Krull's principal ideal theorem. – Slade Jan 29 '15 at 17:30
• Note that $f$ must have positive degree. – Slade Jan 29 '15 at 17:39
• @Slade Can you elaborate on how one can use the Krull's principal ideal theorem? – user26857 Jan 29 '15 at 21:30

There is no need to assume $R$ domain or noetherian.
We have that $R\subset R[X]/(f)$ is an integral extension, so $\dim R[X]/(f)=\dim R$. (Of course, we suppose $\deg f\ge 1$.)
• @user198182 $R[X]/(f)=R[x]$, where $x$ is the residue class of $X$ modulo $(f)$, and $x$ is integral over $R$. – user26857 Jan 29 '15 at 21:38