Proving the ideal $I=\{f(x)\in\mathbb{Z}[x] | f(0)$ is even$\}$ is not principal in the ring $R=\mathbb{Z}[x]$

Asked May 24 '18 at 16:44

Active May 24 '18 at 16:47

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I want to prove that the Ideal $I=\{f(x)\in\mathbb{Z}[x] | f(0)$ is even$\}$ is not principal in the ring $R=\mathbb{Z}[x]$. I have proved that, in fact $I=(2,x)$

Now, I think that should be enough (why not?), since the ideal I is now generated by 2 elements. However my professor followed saying:

To see that $I$ is not principal, we suppose that $I=(f(x))$ and we get to a contradiction. How do I get to that contradiction?

asked May 24 '18 at 16:44

John Keeper

1,281

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Is there a polynomial which divides both $x$ and $2$ in $\Bbb Z[X]$? – Angina Seng May 24 '18 at 16:46
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"Why not": Saying an ideal is generated by two elements does not imply a priori that it is not generated by one element. – David C. Ullrich May 24 '18 at 16:47

Proving the ideal $I=\{f(x)\in\mathbb{Z}[x] | f(0)$ is even$\}$ is not principal in the ring $R=\mathbb{Z}[x]$

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