Product of ideals is a subset of an intersection of ideals So, if I have two ideals of a ring $R$, let's say $I$ and $J$, then we have to prove this:

I tried to take a look at each product of the finite sum. Let's say $a$ is in $I$ and $b$ in $J$. then for the product $ab$, if any of these two elements is in the intersection then we are done. Otherwise, I don't know how to prove that this product will be in the intersection of these ideals.
 A: Take some element $a \in IJ$, which means we have that $a$ is of the form $a = i_1j_1 + \dots + i_nj_n$. Now $i_k \in I$ for all $1 \leq k \leq n$, such that $i_kj_k \in I$ and therefore $a = i_1j_1 + \dots i_nj_n \in I$. Analogously we get $a \in J$. Thus $IJ \subset I \cap J$.
A: If $a\in I$ and $b\in J$, then $ab\in I\cap J$ because both $I$ and $J$ are ideals. 
A: I assume that both $I$ and $J$ are two-sided ideals of $R$.
By definition, 
$IJ = \left \{ \displaystyle \sum_1^n i_k j_k \mid i_k \in I, \; j_k \in J, n \in \Bbb N, \; 1 \le k \le n \right \}; \tag 1$
that is, $IJ$ is the set of finite sums of products of the form $ij$ where $i \in I$, $j \in J$; it is easily seen that $IJ$ is a two-sided ideal in $R$.
Furthermore, since $i_k \in I$, $j_k \in J$ with $I$ and $J$ two-sided, each product 
$i_kj_k \in I, J \Longrightarrow \displaystyle \sum_1^n i_kj_j \in I, J \Longrightarrow  \sum_1^n i_kj_j \in I \cap J;  \tag 2$
now in accord with (2), since $IJ$ is generated by these $i_kj_k$, it follows that
$IJ \subset I \cap J, \tag 3$
$OE\Delta$.
