Let $R$ be an arbitrary ring, $I$ and $J$ be the right and left ideals, respectively. Prove that $IJ \subseteq I \cap J$.
My attempt:
Let $x \in IJ$. That is, $x = i_1j_1+i_2j_2+\dots + i_nj_n$, for all $i_k \in I, j_k \in J, 1\leq k \leq n, n \in \mathbb{N}$. Now, if $i_k \in I, \sum_{k=1}^{n} i_kj_k \in I$ and if $j_k \in J, \sum_{k=1}^{n} i_kj_k \in J$. Hence, $x = I \cap J \Leftrightarrow IJ \subseteq I \cap J$.
As above, may I assume that if $i_k \in I, j_k \in R$ and if $j_k \in J, i_k \in R$ ?
Please correct me if I got mistake way in proving. Thanks in advanced.