Example of an ideal which is not principal in the ring $\mathbb{Z} [x]$

Question

Give an example of an ideal in the ring $\mathbb{Z} [x]$ is not principal. What kind of example would be the easiest?

Look for something generated by two low degree polynomials. Try something really simple, like the constant polynomial $a$ and the polynomial $bx$. — almagest, May 16 '16 at 19:31

score 0 · Accepted Answer · answered May 16 '16 at 20:05

0

Try the following one: $\;I=\langle 2,x\rangle\le\Bbb Z[x]\;$ .

Try first to characterize the elements of the ideal (take a good peek at the free coefficient of this

ideal's elements...), and now assume that $\;I=\langle f(x)\rangle\;,\;\;f(x)\in\Bbb Z[x]\;$ and get a contradiction.

answered May 16 '16 at 20:05

DonAntonio

@user26857 Thank you, both for the downvote and the sarcasm. I don't think it is an answer until he works quite a bit to prove several things. I think many here already forgot how hard can it be to come up with examples for this kind of things when taking a basic undergraduate course. I also think that this site's purpose is not as clear as it declares itself for many. – DonAntonio May 16 '16 at 20:08
You can also read this: http://math.stackexchange.com/questions/1559901/proof-verification-langle-2-x-rangle-is-a-prime-not-principal-ideal – user26857 May 16 '16 at 20:09
And this: http://math.stackexchange.com/questions/629950/why-i-left-px-in-mathbbz-leftx-right2-mid-p0-right-is-not-a-prin – user26857 May 16 '16 at 20:09
@user26857 Thank you. I'd rather help directly someone with something I know than waste my time looking for questions. I've tried that many times and it can be frustrating since it all depends on the first words on the question's header written by the asker. – DonAntonio May 16 '16 at 20:13

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