Proving $f(X)$ is a constant if $f(X)g(X)=1$ Question:
Let $\Bbb{F}$ be a field
Let $f(X), g(X) \in \Bbb F[X]$
Suppose that $f(X)g(X) = 1$. Prove that $f(X)=a$ for some $a \in \Bbb F$
My attempt:
If we assume $\deg f(X)$, $\deg g(X)$ $\ge 0$ then we know 
$\deg f(X)$ $\deg g(X)$ = $\deg f(X)$ + $\deg g(X)$ = 0
Therefore $\deg f(X)= 0 \Rightarrow f(X)=a$ where $a\in \Bbb F$
Would this be a correct proof? Am I allowed to assume that $\deg f(X)$, $\deg g(X)$ $\ge 0$
 A: You proof is fine.
You might just start with the observation that $fg=1$ implies $f\ne0$ and $g\ne0$ and so you can take the degree of both.
Also, it is clearer if you write
$$
\deg f + \deg g = \deg (fg) = \deg 1 = 0
$$
A: One has $\deg(PQ)=\deg(P)+\deg(Q)$ for all $P,Q\in F[X]$, provided one defines $\deg(0)=-\infty$ and $-\infty+a=-\infty$ for all $a\in\{-\infty\}\cup\Bbb N$. Then $PQ=1$ gives $\deg(P)+\deg(Q)=0$, and this implies $\deg(P)=\deg(Q)=0$, and in particular that $\deg(P)\neq-\infty$ and $\deg(P)\neq-\infty$ (or you can say directly that $P=0$ and $Q=0$ are excluded by the condition $PQ=1$). So yes, you can conclude that invertible elements in $F[X]$ are constant polynomials; indeed nonzero constant polynomials.
A: In order to not using any further information, write ($n>0$ integer) $$f(x)=\sum_{k=0}^na_kx^k\;\;\;\text{ and }\;\;\; g(x)=\sum_{l=0}^mb_lx^l$$ where $a_n,b_m\not = 0$. Then $$0=f(x)g(x)-1=a_nb_mx^{m+n}+\sum_{k=1}^{m+n-1}c_kx^k+(a_0b_0-1)$$ and if $R$ is an integral domain (a bit more general case of Field), then $a_nb_m\ne 0$ too. But this implies all coefficients are zero, in particular $$a_nb_m=0$$ which gives a contradiction. (Moreover, $f$ can not be zero too).
