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 $\text{deg}(f(x))=0$
 A: By definition if $f(x)$ is invertible, then there exists a polynomial $g(x)$ such that 

$f(x)g(x)=1$.

Now the degree of the polynomial $1$ is $0$. What can you conclude about the degree of $f$ and $g$?
A: Go for a contradiction.  Assume that $\deg(f)\geq 1$  Write out the polynomial, which has nonzero leading coefficient $a_n$ 
$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$
Write the inverse function with nonzero lead coefficient
$$f^{-1}(x)=b_mx^m+b_{m-1}x^{m-1}+...+b_1x+b_0$$ 
$f(x)f^{-1}(x)$ has the lead term $a_nb_mx^{n+m}$ where $\deg(f(x)f^{-1}(x))= n+m\geq1$.  because $a_nb_m\neq0$ since we are in an integral domain.  Also, $f(x)f^{-1}(x)=1$, which has degree 0.  Contradication.  Therefore, $\deg(f)=0$ if $f$ has an inverse.
A: 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 $0$. 
