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.

  • $\begingroup$ Do you mean Krull dimension? $\endgroup$ – arsmath Jan 29 '15 at 17:27
  • $\begingroup$ @arsmath yes. edited. $\endgroup$ – user198182 Jan 29 '15 at 17:28
  • $\begingroup$ Krull's principal ideal theorem. $\endgroup$ – Slade Jan 29 '15 at 17:30
  • $\begingroup$ Note that $f$ must have positive degree. $\endgroup$ – Slade Jan 29 '15 at 17:39
  • $\begingroup$ @Slade Can you elaborate on how one can use the Krull's principal ideal theorem? $\endgroup$ – 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 $\deg f\ge 1$.)

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  • $\begingroup$ How is it obvious that it is an integral extension? $\endgroup$ – user198182 Jan 29 '15 at 21:37
  • 1
    $\begingroup$ @user198182 $R[X]/(f)=R[x]$, where $x$ is the residue class of $X$ modulo $(f)$, and $x$ is integral over $R$. $\endgroup$ – user26857 Jan 29 '15 at 21:38
  • $\begingroup$ It's clear now. Thank you $\endgroup$ – user198182 Jan 29 '15 at 21:41

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