$R/(IJ)$ is reduced $\Rightarrow IJ = I \cap J$ for ideals $I,J$ of a commutative ring $R$

Question

This is exercise $4.6$ on page $154$ of the textbook Algebra: Chapter $0$ (authored by P. Aluffi):

Let $I,J$ be ideals of a commutative ring $R$. Assume that $R/(IJ)$ is reduced (that is, it has no nonzero nilpotent elements). Prove that $IJ = I \cap J$.

This is how I tried (knowing that $IJ \subseteq I \cap J$ since $ij \in I$ and $ij \in J$, and since $I \cap J$ is an ideal, it is closed under addition):

If $R/(IJ)$ is reduced, then $\forall r \in R: \ r^n \in IJ \Rightarrow r \in IJ$.

Let $a \in I \cap J$. $a^2 = aa \in IJ$, since $a \in I$, but also $a \in J$. Then (since $R/(IJ)$ is reduced) $a \in IJ$.

Is it a right way? If yes, does it mean the commutativity hypothesis is not necessary, and it works for arbitrary rings? If not, where is the mistake?

Answer 1

The proof is correct and the statement doesn't indeed require $R$ is commutative.

What fails in the noncommutative case is that the set of nilpotent elements may not be an ideal, but it is not a problem here.

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You may want to note that a ring has no (nonzero) nilpotent elements if and only if $a^2=0$ implies $a=0$, so the use of just $a^2$ in your proof is not surprising.

Answer 2

Your question is a special case of this theorem.

Theorem: Let $R$ be a reduced ring, and let $I$ and $J$ be two ideals of $R$. If $IJ=0$, then $I\cap J=0$.

Proof: As your proof, since $(I\cap J)^2\subseteq IJ=0$, we have $(I\cap J)^2=0$. Thus, $I\cap J=0$.