5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ How do you define the associated prime ideals ? $\endgroup$ – user18119 Jun 17 '12 at 21:21
  • $\begingroup$ 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. $\endgroup$ – user26857 Jun 18 '12 at 22:38
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.