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Let $\phi :(R,m) \rightarrow (S,n)$ be a local homomorphism of local Cohen-Macaulay rings, where $S$ is a finite $R$-module.

In their proof of Theorem 3.3.7, Bruns&Herzog write that $\dim S = \dim (R/\ker \phi)$.

Question: Where does this equality come from?

Remark: If $\phi$ is a local epimorphism, then it is clear, since $S \cong R / \ker \phi$. So initially i thought that this was a typo. But then the authors give an example right after the proof, in which $\phi$ is not surjective.

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  • $\begingroup$ $S$ is a finite $R$-module, so it is finite over its subring $\phi(R)$. What can you say about the relation between the dimension of $S$ and of that subring? $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '14 at 2:01
  • $\begingroup$ @MarianoSuárez-Alvarez: Since $S$ contains $\phi(R)$ , the annihilator of $S$ as a $\phi(R)$-module is zero. Hence, the Krull dimension of $S$ as a $\phi(R)$-module is the same the Krull dimension of $\phi(R)$. Thanks :) $\endgroup$ – Manos Oct 28 '14 at 2:44
  • $\begingroup$ @MarianoSuárez-Alvarez: But the Krull dimension of $S$ as an $R$-module has nothing to do with the Krull dimension of $S$ as a ring, right? $\endgroup$ – Manos Oct 28 '14 at 2:45
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    $\begingroup$ Since S is a finite module over the image, it is integral over that subring. $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '14 at 3:26
  • $\begingroup$ @MarianoSuárez-Alvarez: You nailed it. If you would like to put these comments in an answer i will gladly accept it. $\endgroup$ – Manos Oct 28 '14 at 3:29
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$S$, being a finite module over $R$, is integral over the image of $\phi$, so has the same dimension as this last ring.

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