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I'm studying on this book

http://books.google.co.in/books?id=ouCysVw20GAC&printsec=frontcover&hl=it#v=onepage&q&f=false

on page 10 there is a Rees Theorem. I'd like to know why the theorem have the hypothesis $IM\neq M$.

Some lines next we have that $IM=M$ implies $\mathrm{Supp}\;M\cap\mathrm{Supp}\;R/I=\emptyset$, why?

I give the statement of the theorem (if you can't use google books the book is Cohen-Macaulay rings of Bruns-Herzog):

Let $R$ be a Noetherian ring, $M$ a finite $R$-module, and $I$ an ideal such that $IM\neq M$. Then all maximal $M$-sequences in $I$ have the same length $n$ given by $n=\mathrm{min}\{i:\mathrm{Ext}^i_R(R/I,M)\neq0\}$.

I believe that all of this is also related to another thing I don't understand, at page 11 the proposition 1.2.10 (a): Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $M$ a finite $R$-module, then $\mathrm{grade}(I,M)=\mathrm{inf}\{\mathrm{depth}\;M_\mathfrak{p}:\mathfrak{p}\in V(I)\}$.

What I don't understand is why $\mathrm{grade}(I,M)=\infty$ implies $\mathrm{depth}\;M_\mathfrak{p}=\infty$.

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    Clic$k$ on the "edited xxxx ago" to see what changes were made in each revision. In this case, I only changed the title to something more helpful and closer to your original title.2011-06-27

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If $\mathfrak p$ is a prime ideal of $M$, then $Ext^i(R/I,M)_{\mathfrak p} = Ext^i(R_{\mathfrak p}/IR_{\mathfrak p},M_{\mathfrak p}).$ In particular, $Ext^i(R/I,M)_{\mathfrak p} = 0$ unless $\mathfrak p \supset I$. On the other hand, if $I M = M$ and $\mathfrak p \supset I$, then $M/\mathfrak p M$ is a quotient of $M/IM$, hence vanishes, and so $M_{\mathfrak p} = 0$.

Consequently, if $IM = M$, then $Ext^i(R/I,M)$ vanishes after localizing at every prime ideal $\mathfrak p$, and so it vanishes (and this is for every $i$). Thus there is no value of $i$ such that $Ext^i(R/I,M) \neq 0$, and so the quantity $n$ is not well-defined in this case.

The above computation also shows that if $\mathfrak p \supset I$ and $IM = M$ then $\mathfrak p \not\in Supp(M)$, i.e. that $Supp(R/I) \cap Supp(M) = \emptyset.$

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    @Jacob: Dear Jacob, This follows from Nakayama's lemma. Regards,2011-06-18