AP + BQ = 1 (Euclidean algorithm)

Question

If $P, Q \in K[X]$ with no roots in common, then there exist $A, B \in K[X]$ such that $AP + BQ = 1 .$ It should be done with the Euclidean algorithm applied on $P$ and $Q$ but I don't know how to do that so that the existence of $A$ and $B$ can be proved.

Answer 1

The claim is false if $K$ is not a field:

Take $P=2$ and $Q=2X$ in $\mathbb Z[X]$. If $AP + BQ = 1$ then $2A(0)=1$, which cannot happen in $\mathbb Z$.

The claim is false if $K$ is not an algebraically closed field:

Take $P=X^2+1$ and $Q=X(X^2+1)$ in $\mathbb R[X]$. If $AP + BQ = 1$ then $(A+BX)(X^2+1)=1$, which cannot happen in $\mathbb R[X]$ since $X^2+1$ is not a unit.

The claim is true if $K$ is an algebraically closed field.

In this case, $P$ and $Q$ have no roots in common iff $\gcd(P,Q)=1$ because the irreducibles in $K[X]$ are exactly the polynomials of degree $1$ when $K$ is an algebraically closed field. The claim then follows from the Extended Euclidean Algorithm.