Most books I have treat primary decomposition only in the Noetherian case. Atyiah-MacDonald goes a step further and prover the uniqueness theorems of primary decomposition without the Noetherian hypothesis. But it seems to me they get a slight different result from the usual one.


**Definitions**


Recall that a prime $P$ is said to be associated to the $A$-module $M$ if there exists $m \in M$ such that $P = Ann(m)$; equivalently $A/P$ injects into $M$. I denote by $Ass(M)$ the set of associated primes. If $A$ and $M$ are Noetherian, this is always not empty.


For an ideal $I$ we let $Ass(I) = Ass(A/I)$. So a prime $P$ is associated to $I$ if and only if $P$ is of the form $(I : x)$ for some $x \in A$.


For the purposes of this question let me say that $P$ belongs to $I$ if and only if $P$ is of the form $\sqrt{(I : x)}$ for some $x \in A$. We call $Bel(I)$ the set of primes belonging to $I$.


Then the result of Atyiah-MacDonald shows that if $I$ has a (minimal) primary decomposition $I = \bigcap Q\_i$, and if we let $P\_i = \sqrt{Q\_i}$, the set of $P\_i$ which appear is exactly $Bel(I)$. The usual formulation gives instead that for $A$ Noetherian this set is $Ass(I)$.


**The problem**


I want to understand the relationship between $Ass(I)$ and $Bel(I)$. Clearly, since prime ideals are radical, $Ass(I) \subset Bel(I)$. In general I see no reason why the opposite inclusion should be true.


Let us see how to go proving the opposite inclusion in a special case. Assume $I$ is decomposable. Then by the result of Atyiah-MacDonald it is enough to show that if we have a minimal primary decomposition $I = \bigcap Q\_i$, and if we let $P\_i = \sqrt{Q\_i}$, then $P\_i \in Ass(I)$.


Let us do this for $P\_1$ and call $R = Q\_2 \cap \cdots \cap Q\_n$. I also call $P = P\_1$, $Q = Q\_1$, so $I = Q \cap R$.


Then observe that $R/I = R/(R \cap Q) \cong (R + Q) /Q \subset A/Q$. Since $Q$ is $P$-primary, $Ass(A/Q) = P$. So $Ass(R/I) \subset \{ P \}$.


If moreover $A$ is Noetherian this set has to be non empty, so $Ass(R/I) = \{ P \}$ and a fortiori it follows that $P \in Ass(A/I)$.


I don't see how to do this without the Noetherian hypothesis, though.


**Questions**



> 
> Is $Ass(I) = Bel(I)$ always, even if $A$ is not Noetherian?
> 
> 
> Is $Ass(I) = Bel(I)$ if we assume that $A$ is not Noetherian, but at least $I$ is decomposable?
> 
> 
> 



Different notions of associated prime (in the non Noetherian case)
Dear Andrea, let $A=K[X\_1,X\_2,\ldots ,X\_n,\ldots]$, the polynomial ring in countably many variables and $I$ be the ideal $I=(X\_1^2, X\_2^2,\ldots )\subset A $. Then


$$\mathcal M=(X\_1,X\_2,\ldots)\in Bel(I) \setminus Ass(I)$$


Indeed, $(I:1)=I$ has as radical $\sqrt I=\mathcal M$, hence $\mathcal M \in Bel(I)$.
But there is no polynomial $x=P(X\_1,X\_2,\ldots ,X\_N)\notin I$ such that $(I:x)=\mathcal M$ because $X\_M$ will not satisfy $X\_M.x\in I$ for $M>N$ [Of course if $x\in I$, we have $(I:x)=A\neq\mathcal M$]


