Let $k$ be a field and $x,y \in K[x,y]$, then $x$ and $y$ are relatively prime.

Question

Let $k$ be a field and $x,y \in K[x,y]$, then $x$ and $y$ are relatively prime. I already proved that 1 is cannot be a linear combination of $x$ and $y$. I have already tried this directly. I supppose there is and $d(x,y) \in k[x,y]$ such $d(x,y) | x$ and $d(x,y)|y$ and if $d'(x,y) \in k[x,y]$ such $d'(x,y) | x$ and $d'(x,y) | y$ then $d'(x,y)|d(x,y)$. How do I prove $d(x,y)=1$ ???

Answer 1

Use the definition of $\gcd$. If $f(x, y)|x$, then $\deg(f)=0$ as a polynomial in $y$. If $f(x, y)|y$, then $\deg(f)=0$ as a polynomial in $x$. The only polynomials with degree $0$ in both $x$ and $y$ are non-zero constant polynomials, and since $K$ is a field, these all are units.

Answer 2

$x$ is prime (by $K[x,y]/x \cong K[y]$ a domain) so $\,\gcd(x,y) = 1$ or $x;\,$ not $x$ else $\,x\mid y\,\overset{\large x\ =\ 0}\Rightarrow\, 0\mid y$