# Embedded associated prime and non zero divisor

$M$ is a finitely generated $A$-module of dimension $d$ such that $G(M)$ is equidimensional and $M$ does not have any embedded prime.

Given $x\in I$, where $I$ is an ideal of $A$, and $\dim G(M)/x^*G(M)<d$ then show that $x$ is non zero divisor of $M$.

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If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours. I voted to close. – Angelo May 21 '13 at 6:29
You could have saved some time by saying " please solve hw problem on page (insert page) of book (insert book name) – aginensky May 21 '13 at 15:45
This is not any homework problem. This is used in Rees theorem of reduction ideals in the book "Joins and intersections" by Hubert Flenner, Leendert J. van Gastel, Wolfgang Vogel. If it seems too easy to you then you could have given some hint for this problem. – riz May 21 '13 at 16:50

Not everybody has the book. So it would be nice if you can explain your notation. Perhaps read how to ask question article.

I believe that your setting is the following: $(R,m)$ is a Noetherian local ring. Let $I \subseteq m$ be an ideal and $M$ a finitely generated $R$-module. Let $G(-)$ denote $gr_I(-)$, and let $x^*$ denote the initial form of $x$ in $G(-)$.

Question: Let $d = \dim R$. Assume that $M$ is failthful and unmixed and $G(M)$ is equidimensional. Let $x \in I$. If $\dim G(M)/x^* G(M) < d$, then $x$ is a non zero-divisor on $M$.

Consider the following exact sequence $$0 \to L \to G(M) \to G(M/xM) \to 0$$ where $L$ is the kernel of the map $G(M) \to G(M/xM)$. It is easy to see that $x^* G(M) \subseteq L$. Hence we have $$0 \to L / x^* G(M) \to G(M)/ x^* G(M) \to G(M/xM) \to 0$$ exact. By the hypothesis the dimension of $G(M)/ x^* G(M)$ is at most $d-1$. This implies that $\dim G(M/xM)$ is at most $d-1$. Since $\dim G(M/xM) = \dim M/xM$, you can conclude $\dim M/xM = d-1$. Since $x$ avoids all the minimal primes ideal of $M$ and $M$ unmixed, $x$ is a non zerodivisor of $M$.

I don't think I used the assumption of $G(M)$ being equidimensional.

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I have already solved the question. If you check that there is 1-1 correspondence between minimal primes of Rees module of $M$ and minimal primes of $M$, then using equidimensionality of $G(M)$ you will get the answer.

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It is great that you have a solution. How do you prove the 1-1 correspondence? For instance, let $R = k[x,y]/(y^2-x^2-x^3)$, $I = (x,y)R$ and $M = R$. Then $R$ has only one minimal prime, but $gr_I(R)$ has two minimal primes unless characteristic of $k$ is $2$. Also, you shouldn't use an answer for a comment.Read the instruction. – Youngsu May 22 '13 at 18:20