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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..

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    $\begingroup$ Mmmm I think that should look for a counter example to show this is not an Ideal $\endgroup$ – Ahlfkushevich Mar 8 at 23:58
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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 $(6,X)$.

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  • $\begingroup$ 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 $X^2 = (2\cdot 0 + X\cdot 1)(3 \cdot 0 + X\cdot 1) \in \Pi$ ? $\endgroup$ – Jos van Nieuwman Mar 9 at 17:31
  • $\begingroup$ @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 $(2\cdot 1 + 0\cdot X)(3\cdot 1 + 0\cdot X)$. $\endgroup$ – jgon Mar 9 at 17:37
  • $\begingroup$ 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. $\endgroup$ – Jos van Nieuwman Mar 9 at 17:47
  • $\begingroup$ is there a nice proof that $(2,X) \cdot (3,X) = (6,X)$? $\endgroup$ – Jos van Nieuwman Mar 9 at 18:30
  • $\begingroup$ @JosvanNieuwman see my edit $\endgroup$ – jgon Mar 9 at 20:48
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$\Pi$ is not an ideal in $\Bbb Z[x]$ because it's not closed under addition: $2x \in \Pi$ and $3x \in \Pi$ but $x \notin \Pi$. $(\Pi) = (6, X).$

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