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It is a homework question which wants us to prove that $I=(x,2)=\{h(x)=xf(x)+2g(x):f,g\in\mathbb Z[x]\}$ is not a principal ideal.

I thought of trying to make a contradiction but I have no idea where to start with.

Could anyone help me with this please?

Andy Z
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    Hint: Suppose it were principle. Then there would be $m(x), n(x), o(x) \in {\mathbb Z}[x]$ such that $m(x)n(x)=x$ and $n(x)o(x) =2$. Look at what $n(x)$ has to look like; its constant term may be particularly englightening. – JoshuaZ Oct 10 '19 at 20:19
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    Start by assume it is principal generated by $a(x)$, say, so $x=b(x)a(x)$ and $2=c(x)a(x)$ for some polynomials $b$ and $c$. Now look at degrees. – Dzoooks Oct 10 '19 at 20:20
  • @JoshuaZ Thanks! – Andy Z Oct 10 '19 at 20:47
  • @Dzoooks Thanks! – Andy Z Oct 10 '19 at 20:47

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