Is the radical of an irreducible ideal irreducible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:48:28Z http://mathoverflow.net/feeds/question/87870 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87870/is-the-radical-of-an-irreducible-ideal-irreducible Is the radical of an irreducible ideal irreducible? Mary 2012-02-08T06:31:28Z 2012-02-13T07:46:52Z <p>I originally posted this to math.stackexchange.com <a href="http://math.stackexchange.com/questions/106223/is-the-radical-of-an-irreducible-ideal-irreducible" rel="nofollow">here</a>. 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.</p> <p>Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is <strong>irreducible</strong> if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$.</p> <p>Question : Assume that $I$ is an irreducible ideal. Must the radical of $I$ be an irreducible ideal?</p> <p>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.</p> http://mathoverflow.net/questions/87870/is-the-radical-of-an-irreducible-ideal-irreducible/88215#88215 Answer by Pham Hung Quy for Is the radical of an irreducible ideal irreducible? Pham Hung Quy 2012-02-11T18:25:57Z 2012-02-13T07:46:52Z <p>I construct a counterexample for your question in the non-noetherian case:</p> <p><strong>(1)</strong> 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 $A$.</p> <p><strong>(2)</strong> 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 $E(k)$</p> <p><strong>(3)</strong> 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).</p> <p><strong>(4)</strong> 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 $R$.</p> <p>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 $R$</p>