A question about degree of a polynomial Let $R$ be a commutative ring with identity $1 \in R$, let $R[x]$ be the ring of polynomials with coefficients in $R$, and let the polynomial $f(x)$ be invertible in $R[x]$. If $R$ is an integral domain, show that $\text{deg}(f(x))=0$

Let $g(x)$ be the inverse of $f(x)$. Then $f(x)g(x) = 1$.
In an integral domain we have
$deg(f(x)g(x)) = deg(f(x)) + deg(g(x))$.
This can be seen straight from the definition of the degree: the leading coefficien of $f(x)g(x)$ is the product of the leading coefficents of $f(x)$ and $g(x)$ so it's nonzero (because $R$ is integral domain). Hence
$deg(f(x)g(x)) \geq deg(f(x)) + deg(g(x))$.
Coefficients of larger powers are zero so we have equality:
$deg(f(x)g(x)) = deg(f(x)) + deg(g(x))$.
Now $deg(f(x)) + deg(g(x)) = deg(1) = 0$ and since $f(x)$ and $g(x)$ are both invertible they can't be the zero polynomial. Hence it must be that both have degree $0$. 
