Show that there exist $a,b \in K [X_1,X_2,\cdots,X_n]$ and $d \in K[X_1,X_2,\cdots,X_{n-1}]$ such that $aF+bG = d.$

Question

Let $K$ be a field. Let $F,G \in K [X_1,X_2,\cdots,X_n]$ be two polynomials which are relatively prime to each other. Show that there exist polynomials $a,b \in K [X_1,X_2,\cdots,X_n]$ and $0 \neq d \in K [X_1,X_2,\cdots,X_{n-1}]$ such that $aF+bG = d.$

How do I prove it? Please help me in this regard.

Thank you very much.

Answer 1

There may be a positive answer in here for $n = 3$. (I can't comment)

Trying to generalize it:

Since $k[x_1,...,x_n]$ is a UFD, $a,b$ as members of $k(x_1,...,x_{n-1})[x_n]$, a PID, have no common factors (Gauss's Lemma). Then you get $af + bg = 1$ and clearing denominators of $f,g$ you have the desired result.