# Proving an ideal isn't principal [closed]

Ok so In lectures we were given an example of an ideal that wasn't principal.

The ideal $(2,x) \lhd \mathbb{Z}[x]$ was shown to not be principal by:

Let $I = (2,x)$ and suppose that $I = (f)$ for some $f \in \mathbb{Z} [x]$. Using Lemma 5.0.3, since $2 \in I$ we know that $2 = fg$ for some $g \in \mathbb{Z} [X]$. It follows that $\text{deg}(f) = 0$ and that either $f = \pm 1$ or $f = \pm 2$. If $f = \pm 1$, then $( f ) = \mathbb{Z} [x]$. This can not be the case if $f = (2,x)$ since $1\notin I$. Similarly $I \neq (\pm 2)$ since $2 + x \in I$, but $2 + x \notin ( 2 )$.

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@user8603: What is your question? – Zev Chonoles Jun 9 '11 at 11:49
@user8603: What is Lemma 5.0.3? – Dennis Gulko Jun 9 '11 at 12:04
I downvoted this question, because no question is asked. User8603, what is the problem you have? – Thomas Rot Jun 9 '11 at 12:12
Sorry I forgot to ask the question. Lemma 5.0.3. is: $( X ) = \left\{ r_{1} a_{1} + ... + r_{n} a_{n}: m \in \mathbb{N}, r_{i} \in R, a_{i} \in X \right\}$ Anyway I understand that the proof is trying to show that the ideal cannot be generated by a single element. My problem is understanding what $(2,x)$ is. – user8603 Jun 9 '11 at 13:02
@user11181: If you sign in as your other account, user8603, you will be able to edit your question to include this (your current posting here is as an answer). The edit button is right below the "algebra" tag. – Zev Chonoles Jun 9 '11 at 13:58
$(2,x)$ is the ideal generated by $2$ and $x$ in $\mathbb{Z}[x]$, i.e. all polynomials of the form $2p(x)+xq(x)$ such that $p(x),q(x)\in\mathbb{Z}[x]$