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.

  • $\begingroup$ So you only are asking about the non-Noetherian case, is that correct? $\endgroup$ – Mahdi Majidi-Zolbanin Feb 8 '12 at 14:24
  • 6
    $\begingroup$ @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 Feb 9 '12 at 1:37

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 $R$

| cite | improve this answer | |
  • 3
    $\begingroup$ Very pretty! I need to remember this example in case I teach commutative algebra. $\endgroup$ – David E Speyer Feb 11 '12 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 Feb 13 '12 at 8:57
  • $\begingroup$ Yes, if I is a primary ideal, then its radical is prime. $\endgroup$ – Pham Hung Quy Feb 13 '12 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 Feb 13 '12 at 16:19
  • 1
    $\begingroup$ 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 Feb 14 '12 at 1:31

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.