Why we can consider both modules as modules over $R_{(p)}$? (Bruns and Herzog, Theorem 1.5.9)

I'm reading Bruns-Herzog's book Cohen Macaulay rings and have a probably elementary question. Why we may consider both modules as modules over $R_{(p)}$ in this theorem? ...

i know that both modules are modules over $R_{(p)}$. but i dont know if depth and $r$ change with changing ring?

The key observation is that $R_p$ and $R_{p^*}$ can be obtained as localizations of $R_{(p)}$. Similarly we can obtain $M_p, M_{p^*}$ as localizations of $M_{(p)}$. So what B&H effectively do in their proof, is to replace $R$ by $R_{(p)}$ and $M$ by $M_{(p)}$. Hence, there is no computation of the depth with respect to a different ring, just a replacement of the underlying ring.
• @user10000: Ok. What we want to do is compute the depth of $M_p$ with respect to $R_p$ (similarly for $M_{p^*}$). How do we obtain $M_p$? We localize $M$ with respect to the prime ideal $p$ of the underlying ring $R$. Now the key point to understand is that $M_{(p)}$ is an $R_{(p)}$-module and we can obtain $M_p$ as a localization of $M_{(p)}$ with respect to the prime ideal $p R_{(p)}$ of $R_{(p)}$. So we can replace the original underlying ring $R$ with $R_{(p)}$ and the original module $M$ with $M_{(p)}$. Is it clear now? Apr 18 '14 at 16:20
• @user10000: The question you are asking does not make sense. You are not proving that the equality holds over $R$ or $R_{(p)}$. You are proving the equality on the localization of these two at $p$, which is the same thing. Apr 19 '14 at 13:47