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Show that the ideal $I=(2,x)$ of $\Bbb Z[x]$ is not principal.

If $I=(2,x)$ is principal, then $(2,x)=(f(x))$ for some $f \in \Bbb Z[x]$. However $(2,x)= \{2p(x) + q(x)x \mid p,q \in \Bbb Z[x] \}$ and since $q(x)x$ has no constant term the constant term of the sum $2p(x) + q(x)x$ will always be even from the contribution of the $2$.

How can I finish the proof from here? I don't know how this contradicts $(2,x)=(f(x))$.

geojaffa
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    Well, if $(2,x)=(f)$ then in particular $f$ divides $2$. What does it tell us about $f$? There are not many options. – Mark Aug 17 '22 at 08:54
  • See https://math.stackexchange.com/questions/500254/is-mathbbzx-a-principal-ideal-domain – Gerry Myerson Aug 17 '22 at 09:02

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