Coprime polynomials ($\gcd (f,g)=1$)

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

Let $u,v\in \mathbb{C}[t]$ be coprime polynomials. If $\alpha u+\beta v$ is a square for four distinct $(\alpha :\beta )$, then $u,v$ are constant. For independent $u,v$, the four square members $L_j=s_j^2$ are pairwise coprime polynomials. Let $d$ and $e$ be their maximal and minimal degrees; at least three have degree $d$. For independent members of degrees $e,d$, the nonzero polynomial $W=L_k^{\prime }{L}_l-L_k{L}_l^{\prime }$ has degree at most $d+e-1$. Each $s_j$ divides $W$, giving $\mathrm{deg}W\ge (3d+e)/2>d+e-1$, a contradiction. The dependent case follows directly from coprimality of polynomials.

For univariate polynomials $f,g$ over a field, not both zero, there are polynomials $A,B$ such that $Af+Bg=d$, where $d$ is their monic greatest common divisor. In particular, coprime polynomials admit $Af+Bg=1$. The extended Euclidean algorithm constructs the coefficients.

For two polynomials over the complex numbers $p,q$ of equal positive degree of a polynomial $n$ with no common root of a polynomial, there are unique polynomials $u,v$ of degree of a polynomial less than $n$ such that $up+vq=1$. To prove the bound, let $V_m$ be the vector space of polynomials of degree of a polynomial at most $m$, including the zero polynomial. The linear map $(a,b)\mapsto ap+bq$ from $V_{n-1}\times V_{n-1}$ to $V_{2n-1}$ has zero kernel of a linear map: if $ap=-bq$, each root of a polynomial of $p$, counted with its multiplicity of a root, must occur in $b$ because it does not occur in $q$, forcing $b=0$ by the degree of a polynomial bound, and then $a=0$. Both vector spaces have dimension $2n$, so the rank-nullity theorem makes the linear map a bijection, and $1$ has a unique preimage.