Is the product IJ of ideals I, J of a ring A an ideal? More specific, Let $A$ be the ring and $I, J$ ideals of $A$.
Then $I\cdot J=$finite sums of elements of the form $u\cdot v$, with $u\in I; v\in J$ is an ideal. Also, I need to prove that $I\cdot J\subseteq I\cap J$.
 A: If $I$ is a left ideal, and $J$ is a right ideal of $A$, then 
$IJ = \left \{ \displaystyle \sum_1^n i_k j_k \mid n \in \Bbb N, i_k \in I, \; j_k \in J \right \} \tag 1$
is a two-sided ideal of $A$.
To prove this, we must establish two facts:
first, that if $a, b \in IJ$, then $a - b \in IJ$; for if $a, b \in IJ$ then we have
$a = \displaystyle \sum_1^{n_a} i_k j_k, \tag 2$
$b = \displaystyle \sum_1^{n_b} i_l j_l, \tag 3$
with $n_a, n_b \in \Bbb N$, $i_k, i_l \in I$, $j_k, j_l \in J$; but 
$a - b = \displaystyle \sum_1^{n_a} i_k j_k - \sum_1^{n_b} i_l j_l \tag 4$
may clearly be written as a sum of the form
$\displaystyle \sum_1^n i_k j_k, \; n \in \Bbb N, \; i_k \in I, \; j_k \in J; \tag 5$
thus
$a - b \in IJ \tag 6$
as it is defined in (1); 
second, that if $a \in IJ$, $r, s \in A$, both $ra, as \in IJ$; since $a \in IJ$ may be written
$a = \displaystyle \sum_1^n i_k j_k, \tag 7$
we have
$ra = \displaystyle \sum_1^n ri_k j_k; \tag 8$
but since $I$ is a left ideal, $ri_k \in I$ for all $i_k \in I$, $r \in A$; thus the sum
$\displaystyle \sum_1^n ri_k j_k \in IJ; \tag 9$
likewise, by a similar argument which exploits the fact that $j_ks \in J$ for $j_k \in J$, $s \in A$,
$as = \displaystyle \sum_1^n i_k j_k s \in IJ. \tag{10}$
We have shown that for $a, b \in IJ$, $r, s \in A$, that
$a - b, ra, as \in IJ, \tag{11}$
and thus $IJ$ satisfies the definition of a two-sided ideal of $A$.
As for 
$IJ \subset I \cap J, \tag{12}$
this is easy to see if we allow $I$, $J$ to be two-sided ideals, since then
$ij \in I, \; ij \in J \Longrightarrow ij \in I \cap J. \tag{13}$
A: Yes it is an ideal. You should check that it is an abelian group. Furthermore, it should be easy to verify that it absorbs multiplication by any element $a \in A.$ As for the last point, note that every element is a finite sum of elements of the form $u \cdot v$ hence every summand of the element is an element of both $I$ and $J$ and hence an element of $I \cap J.$ 
