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Let $A$ be a Noetherian ring and $\mathfrak a$ be an ideal of $A$. Then it is well-known that the associated prime ideas of $\mathfrak a$ are those prime ideals that have the form $(\mathfrak a:x)$ for $x \in A$.

I want to know whether Noetherian condition is necessary or not, that is, for an arbitrary ring $A$ (always commutative with $1$) knowing that an ideal $\mathfrak a$ of $A$ has a minimal primary decomposition, is it possible to obtain the same result?

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How do you define the associated prime ideals ? – user18119 Jun 17 '12 at 21:21
As I already defined in my answer bellow, an associated prime ideal of $I$ must be a minimal prime ideal over an ideal of the form $(I:a)$, $a\in A$. In the Bourbaki terminology this corresponds to the notion of weak associated (faiblement associé) prime ideal. – user26857 Jun 18 '12 at 22:38
up vote 3 down vote accepted

No, it is not! If $A$ is a strongly Laskerian ring, then every associated prime of an ideal $\mathfrak a$, i.e. prime ideal minimal over an ideal of the form $(\mathfrak a:z)$, $z\in A$, has the form $(\mathfrak a:x)$ for some $x\in A$. And, of course, there are strongly Laskerian rings which are not Noetherian.

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