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 $\mathrm{depth}\;M_\mathfrak{p}=\infty$.