Is the radical of the product of ideals equal to the product of the radicals? $ \sqrt{I}\cdot \sqrt{J} = \sqrt{I \cdot J} $?

Question

Question: Is this formula true: $$ \sqrt{I}\cdot \sqrt{J} = \sqrt{I \cdot J} $$ for ideals $I,J$ in a commutative Noetherian ring?

The only result I could find on the internet was this unanswered post on another website.

I know that $$\sqrt{I \cdot J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$$ see here, and here (Lemma 1.7). I want to combine the answer to my question (assuming it is positive), with the aforementioned true result, and the Strong Nullstellensatz, to prove that $Z(I_1) \cup Z(I_2) = Z(I_1 \cdot I_2) = Z(I_1 \cap I_2)$ over an algebraically closed field.

The original motivation for everything.

In $\Bbb Z$, for a prime $p$, we have that $(p)\cdot (p)=(p^2)$ — , Oct 12 '16 at 14:39

score 5 · Accepted Answer · 2016-10-12T15:01:54.793

5

Let $R$ be a noetherian domain which is not a field and $\mathfrak m$ a maximal ideal. By using Nakayama's lemma on the extended ideals in the ring $R_{\mathfrak m} $, we have $\mathfrak m\ne\mathfrak m^2$, therefore $$\sqrt{\mathfrak m}\cdot\sqrt{\mathfrak m}=\mathfrak m\cdot\mathfrak m=\mathfrak m^2\ne \sqrt{\mathfrak m^2}=\mathfrak m$$

edited Oct 12 '16 at 15:01

answered Oct 12 '16 at 14:42

Is the radical of the product of ideals equal to the product of the radicals? $ \sqrt{I}\cdot \sqrt{J} = \sqrt{I \cdot J} $?

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