Let $A$ be a commutative ring with only finitely many minimal prime ideals.

Is the zero ideal $(0)$ decomposable?

[The converse implication is well known. Recall that an ideal is decomposable if it is a finite intersection of primary ideals.]

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    $\begingroup$ @Xam - I find striking the fact that there are only finitely many minimal primes if $(0)$ is decomposable. So it looks natural to me to ask if the converse holds. Also, the only way I know of showing that $(0)$ is not decomposable in a ring $A$ is by proving that $A$ has infinitely many minimal primes. I think the surprise would be if nobody had asked this question before. $\endgroup$ Aug 5, 2017 at 19:07
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    $\begingroup$ I find this question quite interesting, and would like to know if you had any update concerning it. I don't know if it is trivial or not, but I have been thinking about it and the following facts are kind of obvious. Firstly, $A$ is clearly non-Noetherian (otherwise the question is trivial), hence there is no obvious geometry on it. Secondly, because I'm a bit suspicious with maths :) , and perhaps your question in a first glance seems to be correct I was trying to disprove it somehow and ended up looking in the ring of real valued functions $C(X)$, for some space $X$, $\endgroup$
    – user321268
    Aug 19, 2017 at 18:29
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    $\begingroup$ where the above fails to be true. It's kind of trivial to check that this ring is way far from having finitely many minimal primes. So afterwards thought that probably a good quotient of it by a certain ideal would give us the situation, but unfortunately didn't manage to derive something (probably because your question is correct). If you want to give any additional info you extracted after your post, please do collaborate. $\endgroup$
    – user321268
    Aug 19, 2017 at 18:32
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    $\begingroup$ @Pierre-YvesGaillard Please take a look at this example and set $A=R/I$. Then $(0)$, that is $I$ in $R$, has only one a minimal prime $\{0\}\times\mathbb Q$, and $I$ has no primary decomposition. $\endgroup$
    – user26857
    Aug 31, 2017 at 20:33
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    $\begingroup$ @user26857 - Thanks!!! Awesome! If would be great if you could post an answer! I hadn't seen the threads you link to. So my question is an exact duplicate of your first comment to this question... $\endgroup$ Aug 31, 2017 at 23:40


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