$I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.

So I basically have to prove what is on the title. Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $$IJ=I\cap J$$One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.

I am having troubles proving the other inclusion. Any comments?


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Edit: Undeleted and expanded upon as per my comments.

The statement you are trying to prove is only necessarily true for commutative rings. In this case, you can argue that
$$I\cap J= (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$
but the key step $(I\cap J)R$ breaks down in general rngs.

A counterexample to the statement for general rngs is given by endowing the group $\mathbb Z$ with the zero product, that is defining $a\cdot b=0,\forall a,b\in\mathbb Z$. The ideals $(2)$ and $(3)$ are still comaximal, as for any $x\in\mathbb Z$ we can write $x=a\times 2+b\times 3$ for some $a,b\in\mathbb Z$ where $\times$ denotes regular multiplication, and $a\times 2=a+\cdots+a\in(2)$ where addition is performed $a$ times, and similarly $b\times 3\in (3)$. But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.