Let $R$ a commutative ring with unity, and let $I,J\trianglelefteq R$ ideals such that $I+J=R$. Prove that $IJ=I\cap J$. Recall $IJ=\{\sum_{k=1}^n i_kj_k:i_k\in I,j_k\in J, k\in \mathbb{N} \}$.
I have shown that $IJ\subseteq I\cap J$. The rest of the solution does not use the commutativity of $R$ so my question is where is the mistake:
Let $a\in I\cap J$. $a=a\cdot 1\in IJ$ thus $I\cap J=IJ$.
As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $a\cdot 1\in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $a\cdot 1\in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $i\in I,j\in J$. Now if $a\in I\cap J$, try to rewrite $a=a\cdot 1=a\cdot (i+j)=(a\cdot i)+(a\cdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
