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I originally posted this to math.stackexchange.com here. I got a partial answer, but I now suspect that the complete answer is much harder than I thought, so I'm posting it here.

Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is irreducible if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$.

Question : Assume that $I$ is an irreducible ideal. Must the radical of $I$ be an irreducible ideal?

On math.stackchange.com, I learned that the answer is "yes" if $R$ is Noetherian. My guess is that there is a counterexample if $R$ is not assumed to be Noetherian, but I have no idea how to construct it.

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  • $\begingroup$ So you only are asking about the non-Noetherian case, is that correct? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Feb 8, 2012 at 14:24
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    $\begingroup$ @Mahdi Majidi-Zolbanin : I'm interested in the general case, but on math.stackexchange.com someone has already answered the question in the Noetherian case. The non-Noetherian case is all that is left. $\endgroup$
    – Mary
    Commented Feb 9, 2012 at 1:37

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I construct a counterexample for your question in the non-noetherian case:

(1) Let $A = k[[X,Y]]/(XY) = k[[x,y]]$, where $k$ be a field. Notice that $(0) = (x) \cap (y)$ so $(0)$ is reducible in $A$.

(2) We consider the injective hull $E(k)$ of $k$, and set $m \in E(k)$ be the element such that $mA \cong k$. Notice that every non-zero submodule of $E(k)$ contains $mA$ and $(0)$ is irreducible in $E(k)$

(3) Set $R = A \ltimes E(k)$ be the indealization. We have that $(0 \ltimes E(k))^2 = 0$ so $\sqrt{(0)R} = 0 \ltimes E(k)$ is reducible by (1).

(4) We can prove that for every non-zero element $(a,s)$ of $R$, we have $0 \ltimes mA \subseteq (a,s)R$. So the ideal $(0)$ is irreducible in $R$.

EDIT (13/02): It should be noted that this example is also a counterexample for a non-noetherian ring with an ideal is irreducible but not primary. Indeed, we have $(0)$ is irreducible as above. However $$(x,0).(y,m) = (0,0) \in R,$$ and $(x,0)$ and $(y,m)$ are not nilpotent so $(0)$ is not primary in $R$

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    $\begingroup$ Very pretty! I need to remember this example in case I teach commutative algebra. $\endgroup$
    – David E Speyer
    Commented Feb 11, 2012 at 19:54
  • $\begingroup$ Dear Pham Hung Quy: About your edit: It seems to me that if 0 was primary, its radical would be prime, and thus irreducible. $\endgroup$
    – Pierre-Yves Gaillard
    Commented Feb 13, 2012 at 8:57
  • $\begingroup$ Yes, if I is a primary ideal, then its radical is prime. $\endgroup$
    – Pham Hung Quy
    Commented Feb 13, 2012 at 10:53
  • $\begingroup$ Dear Pham Hung Quy: Thanks for having answered my comment. I'm sorry, I still don't see the point of your edit, because I think it is a standard fact that the radical of any primary ideal is irreducible. $\endgroup$
    – Pierre-Yves Gaillard
    Commented Feb 13, 2012 at 16:19
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    $\begingroup$ In my edit: $(0)$ is irreducible but not primary. I think the radical of any primary ideal is prime (any prime ideal is irreducible). In my example $(0)$ is not primary. $\endgroup$
    – Pham Hung Quy
    Commented Feb 14, 2012 at 1:31

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