# Different notions of associated prime (in the non Noetherian case)

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?

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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 $(I:x)=A\neq\mathcal M$]
Thank you very much! Since the radical of $I$ is maximal, $I$ is actually primary, so equality does not hold even in the decomposable case. So I guess this settles everything. :-) – Andrea Ferretti Mar 2 '10 at 21:21