In $\mathbb Z[x]$, show that the only common divisors of $2$ and $x$ are $1$ and $-1$.
$\mathbb Z[x]$ is the ring of poloynomials with integer coefficients.
This should be a pretty trivial question. I'm not really sure how one shows this to be true outside of something glaringly straightforward like:
The divisors of $2$ in $\mathbb Z[x]$ are $1,-1,2,$and $-2$.
The divisors of $x$ in $\mathbb Z[x]$ are $1,-1,x,$ and $-x$.
Thus, the only common divisors are $1$ and $-1$.
Is there something I am missing?
As a side note, the only elements of $\mathbb Z[x]$ that have multiplicative inverses are the nonzero constants, correct?
If $p,q\in\mathbb{Z}[x]$, then $\def\deg{\operatorname{deg}}\deg(pq)=\deg p+\deg q$, with the convention that $\deg 0=-\infty$ and $-\infty+k=-\infty$ for any $k\in\mathbb{N}\cup\{-\infty\}$; $\deg p$ denotes the degree of $p$.
Thus a divisor of $2$ must have degree $0$, so it's a constant, hence we're left with $1$, $-1$, $2$ and $-2$. Now $2$ does not divide $x$, because $2p(x)=x$ means $\deg p=1$ and a contradiction is easily reached. Thus $2$ and $-2$ do not divide $x$, from which $1$ and $-1$ are the only common divisors of $2$ and $x$.