Proving an ideal isn't principal [closed]

Question

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 $2 + x \notin ( 2 )$.

Answer 1

$(2,x)$ 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 $p(x),q(x)\in\mathbb{Z}[x]$