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I'm studying on this book

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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You should include the whole statement of the theorem in the body of your question. Those links to google books don't always allow everybody to see the same pages that you're able to see. – Adrián Barquero Jun 14 '11 at 17:24
"Help on a proof" is a very unhelpful title. Worse, you've now edited two questions into having this exact same title. Please make titles informative, don't edit them to make them less informative. – Arturo Magidin Jun 26 '11 at 23:04
@Arturo: Then next time you will edit a question of mine, please let me know what you edited. – Jacob Fox Jun 27 '11 at 10:49
Click 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. – Arturo Magidin Jun 27 '11 at 15:06

1 Answer 1

up vote 2 down vote accepted

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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I don't understand why $M/\mathfrak{p}M=0$ implies $M_\mathfrak{p}=0$, could you explain me please? – Jacob Fox Jun 18 '11 at 12:08
@Jacob: Dear Jacob, This follows from Nakayama's lemma. Regards, – Matt E Jun 18 '11 at 18:12

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