What does the ideal $I$ of $\mathbb{Z}[X]$ generated by $x$ and $2$ look like?

Question

I don't know how to put it into terms of more explicit set notation

This was marked as a duplicate, but it is NOT. The questions are totally different. Please read the question before marking it as duplicate. — Fredrik, May 03 '16 at 17:51

score 7 · Answer 1 · answered Jan 16 '16 at 00:56

7

$I=(x,2)=\{xf+2g:f,g\in\mathbb{Z}[X]\}$

answered Jan 16 '16 at 00:56

Rafael

1

This is a good answer because it is generalizable (+1). – Matt Samuel Jan 16 '16 at 01:05

score 2 · Accepted Answer · answered Jan 16 '16 at 01:04

2

Every element of $I=(2,x)$ will be exactly the same as any polynomial in $\mathbb{Z}[X]$, except for the constant coefficient being necessarily even.

answered Jan 16 '16 at 01:04

Ottavio

So if I understand you correctly, would it be correct to say that this ideal is principal? – Fredrik Jan 16 '16 at 01:08
1

No, the ideal $(2,x)$ can't be generated by just one polynomial, in $\mathbb{Z}[X]$. Note that only the coefficient of the term of degree zero must be even, any other coefficient can be odd. – Ottavio Jan 16 '16 at 01:15
$(2,x)$ is a notable example of a non-principal ideal, frequently used to show that $\mathbb{Z}[X]$ is not a PID (an integral domain in which every ideal is principal). You can find a proof and further information here: http://math.stackexchange.com/questions/500254/is-mathbbzx-a-principal-ideal-domain – Ottavio Jan 16 '16 at 01:19

2 Answers2