Show that $f(x,y) - f(y,x)$ is divisible by $ y-x \in F[x,y] $

Question

Let $F$ be a field. Show $ f(x,y) -f(y,x)$ is divisible by $y-x \in F[x,y]$ for all $f \in F[x,y]$.

My hope was to use the "evaluation at x" homomorphism $ ev_{y=x} : F[x][y] \rightarrow F[x]$. Clearly $f(x,y) - f(y,x)$ and $y-x$ are both in the kernel, but since $F[x,y]$ is not a PID, I don't think this will help me.

I've been able to show this using brute force, but I was wondering if there were a different approach to this problem.

Answer 1

Hint: $f(x,y)=\sum a_{nm}x^ny^m$ implies that $f(x,y)-f(y,x)=\sum a_{nm}(x^ny^m-x^my^n)$

If $n<m, x^ny^m-x^my^n=x^ny^n(y^{m-n}-x^{m-n})$.

Use $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$

Answer 2

Let $h(x,y) = f(x,y+x) - f(y+x,x)$. Then $h(x,0) = 0$, so that $h(x,y)$ has no terms which only have $x$ and not $y$. Hence $y$ divides each term of $h(x,y)$ and therefore $h(x,y)$ itself. Thus $y - x$ divides $h(x,y-x) = f(x,y) - f(y,x)$.

Answer 3

$\ g(x) := f(x,y)\!-\!f(y,x)\in\!\!\!\!\!\!\overbrace{ R[x]}^{\Large F[y][x]\ \ \ \ }\!\!\!\!\!$ has $\ g(y)=0\ $ so $\, x\!-\!y\mid g(x)\,$ in $\,R[x]\,$ by the Factor Theorem.