Let $R$ denote the set of all finite formal sums of elements in the free group $\left<a,b\right>$ with the relation $a+b=1.$ Let $I$ be the principle ideal generated by $a$ and $J$ the principle ideal generated by $b.$ It follows that $ba\in I\cap J,$ however, $ba\not\in IJ.$
The fact that $ba\not\in IJ$ follows from if $$\sum_{i=1}^mw_i\in I$$ where $w_i$ are words in $a,b,$ then by definition each $w_i$ contains $a.$ Similarly if $$\sum_{j=1}^mv_i\in J,$$ then each $v_i$ contains $b$. Then $$\left(\sum_{i=1}^mw_i\right)\left(\sum_{j=1}^mv_i\right)=\sum_{i,j}w_iv_j,$$ and each $w_iv_j$ has an $a$ to the left of a $b.$ So $\sum_{i,j}w_iv_j\not=ba.$
Clearly $I+J=R$ as $a+b\in I+J$ and $a+b=1.$