Product of ideals in $\mathbb{Z}[X]$

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..

• Mmmm I think that should look for a counter example to show this is not an Ideal – Ahlfkushevich Mar 8 at 23:58

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 $$(6,X)$$.
• Sanity check. This is because $2 = 2 \cdot X^0 + 0\cdot X^1 \in (2,X)$ and $3 = 3 \cdot X^0 + 0\cdot X^1 \in (3,X)$, so $6 = 2\cdot 3 = (2 \cdot 1 + X)(3 \cdot 1 + X) = (2 \cdot X^0 + 0\cdot X^1)\cdot (3 \cdot X^0 + 0\cdot X^1) \in \Pi$. Similarly: $X = (2\cdot 0 + X\cdot 1) = (3\cdot 0 + X\cdot 1)$, so $X^2 = (2\cdot 0 + X\cdot 1)(3 \cdot 0 + X\cdot 1) \in \Pi$ ? – Jos van Nieuwman Mar 9 at 17:31
• @JosvanNieuwman Yes, that's essentially correct, but there appears to be a typo. You wrote $6=2\cdot 3 = (2\cdot 1 + X)(3\cdot 1 + X)$ instead of $(2\cdot 1 + 0\cdot X)(3\cdot 1 + 0\cdot X)$. – jgon Mar 9 at 17:37
• Correct, thanks. I'm now verifying that $6+X^2 \in \Pi \Rightarrow (1) \in \{(2,X), (3,X)\}$. Thanks for the answer. I wouldn't have thought looking for it in not being even a subgroup. I was dabbling only with the multiplication at first. – Jos van Nieuwman Mar 9 at 17:47
• is there a nice proof that $(2,X) \cdot (3,X) = (6,X)$? – Jos van Nieuwman Mar 9 at 18:30
$$\Pi$$ is not an ideal in $$\Bbb Z[x]$$ because it's not closed under addition: $$2x \in \Pi$$ and $$3x \in \Pi$$ but $$x \notin \Pi$$. $$(\Pi) = (6, X).$$