Let $R$ be a commutative ring and let $I$ and $J$ be ideals of $R$. Show $IJ$ is an ideal of $R$. Let $R$ be a commutative ring. 
For ideals $I$, $J \in R$ define $IJ$ to be the set 
$\{a_1b_1 +\ldots+a_nb_n : n\in\mathbb N$; $a_i\in I$; $b_j \in J\}.$ 
Prove that $IJ$ is an ideal in $R$.

Suppose $I, J$ are ideals of a commutative ring $R$. We show $IJ = \{i_1j_1 + i_2j_2 + \cdots + i_nj_n\ : i_k \in I,\, j_k \in J,\, n \in \mathbb Z^+ \}$ is also an ideal.
We must first show that $IJ$ is an additive subgroup of $R$. Observe that $0 \in I$ and $0 \in J$ since each are ideals and therefore additive subgroups of $R$. Thus $0 \cdot 0 = 0 \in IJ \neq \emptyset$. Now let $x \in IJ$. then $x = i_1j_1 + i_2j_2 + \cdots + i_nj_n$ for some $i_k \in I$, $j_k \in J$ and $n \in \mathbb Z^+$. Observe that since $i_k \in I$ and $I$ is an additive subgroup of $R$ by definition of ideal then $-i_k \in I$ and thus 
                \begin{align*}
     (-i_1)j_1 + (-i_2)j_2 + \cdots + (-i_n)j_n &=  -i_1j_1 - i_2j_2 - \cdots - i_nj_n \\
     &= -( i_1j_1 + i_2j_2 + \cdots + i_nj_n) \\
     &= -x \in IJ
    \end{align*}
                Now let $x, y \in IJ$ then $x = i_1j_1 + i_2j_2 + \cdots + i_nj_n$ and $y = i_1'j_1' + i_2'j_2' + \cdots + i_{n'}'j_{n'}'$ for some $i_k, i_k' \in I$, $j_k, j_k' \in J$ and $n, n' \in \mathbb Z^+$. Then $$x + y =\big( i_1j_1 + i_2j_2 + \cdots + i_nj_n\big) + \big( i_1'j_1' + i_2'j_2' + \cdots + i_{n'}'j_{n'}'\big) \in IJ$$ Thus we have shown that $IJ$ is an additive subgroup of $R$.
Now let $r \in R$ and $a \in IJ$ where $i_1j_1 + i_2j_2 + \cdots + i_nj_n$. Observe that $ra = r(i_1j_1 + i_2j_2 + \cdots + i_nj_n) = ri_1j_1 + ri_2j_2 + \cdots + ri_nj_n$ and since $ri_k \in I$ since $I$ is an ideal we an conclude that $ra \in IJ$ and thus $IJ$ is a left ideal.
Since $R$ is commutative $IJ$ is also a right ideal.
This by definition $IJ$ is an ideal of $R$
