Let $I = (2,x)$, an ideal in $\mathbb{Z}[x]$. Show there is an element of $II$ not of the form $ab$, where $a, b \in I$.

Asked May 24 '20 at 07:22

Active May 24 '20 at 07:22

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Let $I = (2,x)$, an ideal in $\mathbb{Z}[x]$. Show there is an element of $II$ not of the form $ab$, where $a, b \in I$.

$II$ is defined as $$I^2 = II = \bigg\{\sum_{i=1}^n r_ia_ib_i \;\bigg\vert\; a_i,b_i \in I, r_i \in \mathbb{Z}[x], n \in \mathbb{N}\bigg\}$$

I know that $I = \{2f(x) + xg(x) \;\vert\; f(x), g(x) \in \mathbb{Z}[x]\}$.

So an element of $I^2$ of the form $ab$ would be $$(2f_1(x) + xg_1(x))(2f_2(x) + xg_2(x)) = 4f_1(x)f_2(x) + 2xf_2(x)g_1(x) + 2xf_1(x)g_2(x) + x^2g_1g_2(x).$$

But I'm stuck on finding out how to show there's something in $I^2$ that doesn't have this form. Any hints?

asked May 24 '20 at 07:22

user594147

How about $4+x^2$? – user350031 May 24 '20 at 07:27
@user350031 I can see why it's not of the form $ab$, but could you explain how you would approach a question like this? – user594147 May 24 '20 at 07:39
The problem with defining the ideal product $IJ$ to be just the set of products $ab$ where $a \in I$ and $b \in J$ is that this is not necessarily closed under addition so this gave me the hint to take a sum of two things of the form $ab.$ Does that help? :) – user350031 May 24 '20 at 07:43
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Also $4+x^2$ is irreducible in $\mathbb{Z}[x]$ so I know that it can't be the product of two things in $(2,x).$ – user350031 May 24 '20 at 07:44
@user350031 Yes, thanks very much! – user594147 May 24 '20 at 18:11
More generally see here. – Bill Dubuque May 24 '20 at 18:19

Let $I = (2,x)$, an ideal in $\mathbb{Z}[x]$. Show there is an element of $II$ not of the form $ab$, where $a, b \in I$.

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