Given that $R$ is a commutative ring and $I,J$ are ideals in $R$, prove that $(I+J)(I\cap J) \subset IJ.$ As stated in the title, it is given that $R$ is a commutative ring and $I,J$ are ideals in $R$. I want to prove that the following is true:
$$(I+J)(I\cap J) \subset I J.$$
I have proven that both $I+J$ and $I\cap J$ are ideals, but not sure how to proceed further.
Any help would be appreciated.
 A: I'll use $i$s to represent elements of $I$ and $j$s to represent elements of $J$.  
Suppose that $a \in I \cap J$, and $b \in I + J$.  Then we can write $b = i_1 +j_1$ and $a = i_2 = j_2$.  Thus $ab = ai_1 + aj_1 = j_2i_1 + i_2j_1 \in IJ$.  
Any element of $(I \cap J)(I+J)$ is thus a finite sum of elements in $IJ$, so being an ideal it is contained in $IJ$.  
Note that this observation is usually used as the first step in the proof of the Chinese Remainder Theorem.  We have just seen that $(I+J)(I \cap J) \subset IJ \subset I \cap J$ (the last inclusion is trivial).
Thus if $I$ and $J$ are comaximal, that is, $I +J = R$, then we have that $IJ = I \cap J$.  
A: It's not difficult to show that, for ideals $A,B,C$, $(A+B)C=AC+BC$. Thus
$$
(I+J)(I\cap J)=I(I\cap J)+J(I\cap J)
$$
and all it's needed is to show that
$$
I(I\cap J)\subseteq IJ
\qquad\text{and}\qquad
(I\cap J)J\subseteq IJ
$$
Can you do it? There are some inclusion relations to be checked.
A: $I + J = \{ i + j \mid i \in I, j \in J \}; \tag 1$
$I \cap J = \{k \mid k \in I, k \in J \}; \tag 2$
for any two ideals $L$ and $K$,
$LK = \displaystyle \left \{ \sum_{p = 1}^n l_p k_p \mid n \in \Bbb N, l_p \in L, k_p \in K  \right \}; \tag 3$
thus
$(I + J)(I \cap J) = \displaystyle \left \{ \sum_{p = 1}^n (i_p + j_p) \cdot k_p \mid n \in \Bbb N, i_p \in I, j_p \in J, k_p \in I \cap J \right \}; \tag 4$
we observe that the sum
$\displaystyle \sum_{p = 1}^n (i_p + j_p)k_p = \sum_{p = 1}^n i_p k_p + \sum_{p = 1}^n j_p k_p; \tag 5$
since $i_p \in I$ and $k_p \in I \cap J \subset J$, 
$\displaystyle \sum_{p = 1}^n i_p k_p \in IJ; \tag 6$
as for the second sum on the right of (5), we note that $k_p \in I \cap J \subset I$, and then by the commutativity of $R$,
$\displaystyle \sum_{p = 1}^n j_p k_p = \sum_{p = 1}^n k_p j_p \in IJ; \tag 7$
it therefore follows that
$\displaystyle \sum_{p = 1}^n i_p k_p + \sum_{p = 1}^n j_p k_p = \sum_{p = 1}^n k_p j_p \in IJ; \tag 8$
we thus see that
$\displaystyle \left \{ \sum_{p = 1}^n (i_p + j_p) \cdot k_p \mid n \in \Bbb N, i_p \in I, j_p \in J, k_p \in I \cap J \right \} \subset IJ; \tag 9$
finally, returning to (4) we conclude
$(I + J)(I \cap J) \subset IJ, \tag{10}$
as required.
