If $I=(a_1,\dots, a_m)$ and $J=(b_1, \dots, b_n)$ are ideals in a commutative ring, then we have
\[IJ=(a_ib_j),\]
where $1\leq i \leq m$ and $1\leq j \leq n$.

Proof.

(a) Prove that $IJ=(x, 6)$.

Note that the product ideal $IJ$ is generated by the products of generators of $I$ and $J$, that is, $x^2, 2x, 3x, 6$. That is, $IJ=(x^2, 2x, 3x, 6)$.

It follows that $IJ$ contains $x=3x-2x$ as well. As the first three generators can be generated by $x$, we deduce that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

Assume that $x=f(x)g(x)$ for some $f(x)\in I$ and $g(x) \in J$.
As $\Z[x]$ is a UFD, we have either
\[f(x)=\pm x, g(x)=\pm 1, \text{ or } f(x)=\pm 1, g(x)=\pm x.\]

In the former case, we have $1\in J$ and hence $J=\Z[x]$, which is a contradiction.
Similarly, in the latter case, we have $1\in I$ and hence $I=\Z[x]$, which is a contradiction.
Thus, in either case, we reached a contradiction.

Hence, $x$ cannot be written as the product of elements in $I$ and $J$.

Comment.

Let $I$ and $J$ be an ideal of a commutative ring $R$.
Then the product of ideals $I$ and $J$ is defined to be
\[IJ:=\{\sum_{i=1}^k a_i b_i \mid a_i\in I, b_i\in J, k\in \N\}.\]

The above problems shows that in general, there are elements in the product $IJ$ that cannot be expressed simply as $ab$ for $a\in I$ and $b\in J$.

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