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An ideal $I$ of a commutative unital ring $R$ is called decomposable if it has a primary decomposition.

Can you give me an example of an ideal that is not decomposable? All the examples I can think of are decomposable. Thanks.

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any primary ideal is decomposable namely itself, I think you want an example of an ideal (not primary ) which is not decomposable. –  pritam Jun 30 '12 at 11:47
    
@pritam Right, thank you! –  Matt N. Jun 30 '12 at 11:52

1 Answer 1

up vote 4 down vote accepted

Zero ideal in $C[0,1]$ is not decomposable.

More generally,

If $X$ is an infinite compact Hausdorff space then the zero ideal of $C(X$), the ring of real valued continuous functions on $X$ is not decomposable.

Proof : First let us note that every maximal ideal of $C(X)$ is of the form $M_x=\lbrace f \in C(X) : f(x)=0\rbrace$, for some $x \in X$. Note that, if the zero ideal of $C(X)$ were decomposable, then there would be only finitely many minimal prime ideals of C(X). This certainly sounds strange since every maximal ideal $M_x$ of $C(X)$ contains a minimal prime ideal as every maximal ideal is also prime ideal. Hence to show that the zero ideal of $C(X)$ is not decomposable it is enough to show that if $x\not= y\in X$ then any two minimal prime ideals $P_1\subseteq M_x, P_2\subseteq M_y$ are different: Since $X$ is Hausdorff and normal, there is an open set $U$ such that $x\in U$ and $y\not\in\overline{U}$. By Urysohn's lemma there are$ f,g\in C(X)$ such that $f(U) = 0, f(y) = 1, g(x) = 1,$ and $g(X \setminus U) = 0.$ So $fg = 0$. Therefore $fg \in P_1$ but $g \not\in P_1$, because $g(x)\not=0$, hence $f \in P_1$. But $f \not\in P_2$, because $f(y)\not=0$. Hence $P_1\not=P_2$.

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Perfect, pritam! +1. –  Georges Elencwajg Jun 30 '12 at 13:10
    
I just saw that this is exercise 6 on page 55 (Ch. 4) in Atiyah-MacDonald. –  Matt N. Jul 1 '12 at 8:40

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