Let $F,G\in\mathbb{F}\left[X\right]$ be polynomials such that $\deg F=n,\deg G=m$. And $\gcd(F,G) = 1$ Then exists $A,B\in\mathbb{F}\left[X\right]$ such that $\deg A\leq m-1,\deg B\leq n-1$ and $AF+GB=1$
I know that from Bézout's polynomial remainder theorem $A,B$ do exist, But I don't know how to show that there can be $A,B$ such that $\deg A\leq m-1,\deg B\leq n-1$. Any ideas on how to approach this?
Suppose you have found polynomials $C, D$ such that $$ C F + D G = 1. $$ Divide $C$ by $G$, to obtain $$ C = Q G + R, $$ for some polynomials $Q, R$, with $R$ being either zero, or nonzero, of degree less than $m$. Then $$ 1 = C F + D G = (Q G + R) F + D G = R F + (D + Q F) G. $$ Note first of all that if $R = 0$, then $G$ is a nonzero constant, so $m = 0$, and we can simply choose $A = 0$ and $B = G^{-1}$.
If $R \ne 0$, then the degree of $1 - R F$ is greater than $0$, and at most $(m-1) + n$. But then so is the degree of $(D + Q F) G$. Since $G$ has degree $m$, it follows that $D + Q F$ has degree at most $n - 1$.
Then you can choose $A = R$ and $B = D + Q F$.
