This might be a little bit too precise and too long question, but I really don't have any where else to turn. The thing is that I want to prove the first and second uniqueness theorems for primary decomposition of finitely generated modules over a Noetherian ring USING the fact that I already proven the corresponding theorems for ideals of Noetherian Rings.

I will just state the things I know in a compact form:

I have a primary decomposition of the submodule $N$ of the $R$-module $M$. Where $N=\bigcap_{i=1}^n N_i$ and $rad(N_i)=rad(N_i:M)=rad(ann(M/N_i))=P_i$. $P_i$ prime ideal of $R$ and $(N_i:M)$ is a $P_i-primary$ ideal of $R$. I know that $(N:M)$ is an ideal of $R$ and since $(N:M)=(\bigcap_{i=1}^n N_i:M)=\bigcap_{i=1}^n(N_i:M)$ this is a primary decomposition of $(N:M)$ and hence the associated primes of $(N:M)=\{P_1,P_2,...,P_n\}$ by the first uniqueness theorem on primary decompositions of ideals. I want to somehow show that $Ass(M/N)=\{P\in Spec(R) :P=ann(x), x\in M/N\}$ is equal to this set of Associated primes to $(N:M)$ which are (the so called associated primes to an ideal I is really Ass(R/I)) $Ass(R/(N:M))=\{P\in Spec(R) :P=ann(x), x\in R/(M:N)\}$.

I guess what I try to deduce is that $(x:(N:M))=(x:N)$. All comments are extremely welcome. Concerning this last equality or the argument stating that the same is a sufficent equality to show. Thanks for an amazing site.


1 Answer 1


In general $\mathrm{Ass}(M/N)=\mathrm{Ass}(R/(N:M))$ is false as shows the following example: $R=\mathbb{Z}$, $M=\mathbb{Z}\times\mathbb{Z}_6$ and $N=0$.


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