Show that 2 and $x$ are coprime in $\mathbb{Z}[x]$

Question

I want to show that $2$ and $x$ are coprime in the ring $\mathbb{Z}[x]$. I realize this is not a PID precisely because of the ideal generated by these two elements-- does this mean I can't show that they generate the unit ideal? If so how would I show this?

Answer 1

Hint $\ x\,$ is prime by $\,\Bbb Z[x]/(x) \cong \Bbb Z\,$ is a domain, and $\,x\nmid 2,\,$ else $\ x f(x)=2\,\overset{\large x\,=\,0}\Rightarrow\, 0 = 2\,$ in $\Bbb Z$

Answer 2

Well, there are also other nonprincipal ideal, but yes, the ideal $(2,x)$ is not principal. Also, it is a proper ideal, because for example $1\notin (2,x)$. In fact you can show that $(2,x)=\{f(x)\in \mathbb Z[x]\colon f(0)\equiv 0 \bmod 2\}$. Showing that $2$ and $x$ are coprime is very easy, since every divisor of $2$ must have degree $0$, and is therefore a constant, while $x$ has no divisors of degree $0$ different from $\pm 1$. So note that being coprime does not imply satisfying Bezout's identity, for which you need to be in a Euclidean domain (I believe).