Consider the ideals $I = (2,X), J = (3,X) \in \mathbb{Z}[X]$. I want to show that the 'product set' $\Pi := \{ij \mid (i,j) \in I \times J\}$ is not an ideal in $\mathbb{Z}[X]$ and in particular, unequal to $IJ = (\Pi)$.
I'm having a hard time getting 'a feel' for this product (i.e. the set generated by $\Pi$). I've been writing out some expressions, but they aren't pretty.
Note that \begin{align*} (2,X) &= 2\mathbb{Z}[X] + X\mathbb{Z}[X] = \{2p + Xq \mid p,q \in \mathbb{Z}[X]\} \\ &= \left\{2\sum_{i=0}^n a_iX^i + X\sum_{i=0}^m b_iX^i \,\,\middle| \,\,\sum_{i=0}^n a_iX^i , \sum_{i=0}^m b_iX^i \in \mathbb{Z}[X]\right\} \\ &= \left\{\sum_{i=0}^n (2a_i + Xb_i)X^i \,\,\middle| \,\,...\right\} \\ \end{align*} and similarly: $\,\,\,(3,X) = \left\{\sum_{i=0}^n (3c_i + Xd_i)X^i \,\,\middle| \,\,...\right\}$.
So any $p \in \Pi$ is of the form $$\sum_{i=0}^{2n}\sum_{j=0}^i (6a_jc_{i-j} + (2d_{i-j}+3b_j)X + b_jd_{i-j}X^2)X^i.$$
Now $(\Pi)$ contains all elements generated by these dragons, so I'm not really getting the sense I'm going about this right...
Edit
I'm computing the product ideal of $(2,X)$ and $(3,X)$, giving me
\begin{align*} (2,X)\cdot (3,X) &= (\{i\cdot j \mid i \in (2,X), j\in (3,X)\}) \\ &= (\{(2p(X) + Xq(X))(3r(X)+ Xs(X)) \mid p(X), q(X), r(X), s(X) \in \mathbb{Z}[X]\}) \\ &= (\{6p(X) + X(2p(X)s(X) + 3q(X)r(X)) + X^2q(X)s(X) \mid p(X), q(X), r(X), s(X) \in \mathbb{Z}[X]\}) \end{align*}
This tells me that all coefficients of the constant polynomials in this product ideal must divide 6. The actual proof that this equals $(6,X)$ still eludes me though..
The product ideal contains $6$ and $X^2$, so it contains $6+X^2$, but the product set cannot contain this element, since $6+X^2$ is irreducible, so if $6+X^2=ij$ for $i\in I$, $j\in J$, we would have one of $i$ or $j$ is a unit, but neither ideal is the unit ideal.
Edit: To see that $(2,X)(3,X)=(6,X)$, note that the product is $(6,2X,3X,X^2)$, and this ideal equals $(6,X)$.