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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 $R$.

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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 $R$

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