On the statement of Theorem 3.3.7 in Bruns and Herzog

Question

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.

Answer 1

$S$, being a finite module over $R$, is integral over the image of $\phi$, so has the same dimension as this last ring.