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Give an example of a commutative ring with unit and an ideal that has no primary decomposition.

I think boolean Ring will be the right example, but I don't know how I must show that. So please help me.

Boolean ring is $R=P(\mathbb{N})$ (the power set of $\mathbb{N}$) with $$A+B=(A\cup B)-(A\cap B)$$ $$AB=A\cap B$$

Also if you have another example, please tell me, thank you.

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  • $\begingroup$ Already given in this topic. $\endgroup$
    – user26857
    Dec 19, 2013 at 21:15
  • $\begingroup$ I mean that any Ideal of a Ring has no primary decomposition. $\endgroup$
    – kpax
    Dec 20, 2013 at 5:33
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    $\begingroup$ @kpax Let's think a little before asking: A commutative ring with unit has (at least) a maximal ideal. Every maximal ideal is prime, and every prime is primary, hence has a primary decomposition. The conclusion: there is no commutative ring with unit such that any ideal has no primary decomposition. $\endgroup$
    – user26857
    Dec 20, 2013 at 6:23
  • $\begingroup$ thank you very much,you make my mind in the right direction. $\endgroup$
    – kpax
    Dec 20, 2013 at 8:11
  • $\begingroup$ Dear @YACP, I see I repeated that argument of yours about maximal ideals (in a comment I made to my answer) without acknowledging you. Sorry about that: I have removed my redundant comment . $\endgroup$
    – Georges Elencwajg
    Dec 20, 2013 at 12:29

1 Answer 1

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Let $X$ be a compact Hausdorff infinite topological space and $C(X)$ the ring of continuous real functions $X\to \mathbb R$.
Then in that ring the zero ideal does not have a primary decomposition.
Indeed, if that were the case $C(X)$ would only have finitely many minimal prime ideals.
But actually $C(X)$ has infinitely many minimal prime ideals because:

a) The maximal ideals consist of the ideals $\mathfrak m_x$ of functions vanishing at $x\in X$ and there are thus infinitely many.
b) every maximal ideal contains at least one minimal prime $\mathfrak p_x \subset \mathfrak m_x$.
c) Whichever choice of the $\mathfrak p_x $'s was made in b) we automatically have $\mathfrak p_x\neq \mathfrak p_y$ for $x\neq y$.
[Use a function $f$ with $f\equiv 0$ near $x$, $f(y)=1$ and a function $g$ with $g(x)=1$ and $g\equiv 0$ near $y$ such that $fg=0$.]

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    $\begingroup$ This is a solution to Exercise 6 of Chapter 4 in Atiyah-Macdonald. $\endgroup$
    – Georges Elencwajg
    Dec 19, 2013 at 20:15
  • $\begingroup$ thank you for giving the reference,but I think that,I need a Ring which any Ideal of it, has no primary decomposition,It was the thought that comes to my mind.if I'm wrong please explain to me. $\endgroup$
    – kpax
    Dec 20, 2013 at 5:38
  • $\begingroup$ I understand,thank you very much for your answer.sorry because of delay in responding,I had internet connection problem. $\endgroup$
    – kpax
    Dec 20, 2013 at 8:16
  • $\begingroup$ Dear @kpax, I'm glad I could help you and I find that you responded quite fast ! (Anyway you were under no obligation to do so.) $\endgroup$
    – Georges Elencwajg
    Dec 20, 2013 at 8:28
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    $\begingroup$ @kpax - You may already have discoverd this on your own, or, perhaps you are not interested in the problem anymore, but notice that the primes associated to a primary decomposition of $(0)$ are finite in number. The minimal members of this set are precisely the set of minimal elements in the set of all primes containing $(0).$ See the discussion surrounding Theorems 4.5 and 4.6 of Atiyah-Macdonald. So, $(0)$ decomposable implies that the ring in question has only finitely many minimal prime ideals. $\endgroup$
    – Chris Leary
    Jan 26, 2018 at 20:49

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