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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.

Aydin Ozbek
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    For any commutative ring $A$, the map of evaluation at $a \in A$, $ev_a : A[t] \to A$, has kernel $\langle t - a \rangle$. Now use this in the case $A = F[x]$ and $a = x$. – Daniel Schepler Jun 20 '19 at 23:14
  • @DanielSchepler I am trying to prove the fact you stated, but I am having trouble showing that $f \in \ker ev_a $ implies that $ f = (t-a)f' $ as there is no division algorithm for polynomials in a commutative ring. Do you know how to show $f \in (t-a) $? – Aydin Ozbek Jun 21 '19 at 00:37
  • You do still have a division algorithm for dividing by monic polynomials with coefficients in a general commutative ring. – Daniel Schepler Jun 21 '19 at 01:06
  • Alternatively, for this specific case, a generalization of Horner's method for dividing by $t-a$ will still work: if $p(t) = c_n t^n + c_{n-1} t^{n-1} + \cdots + c_0$, and you form $b_n = c_n$, $b_{n-1} = b_n a + c_{n-1}$, $b_{n-2} = b_{n-1} a + c_{n-2}$, $\ldots$ then $p(t) = (b_n t^{n-1} + b_{n-1} t^{n-2} + \cdots + b_1) (t-a) + b_0$. – Daniel Schepler Jun 21 '19 at 01:12
  • The remark of @Daniel about the universality of division by monic polynomials is proved here, which implies that the Factor Theorem works universally, which immediately yields the claim - see my answer. – Bill Dubuque Jun 21 '19 at 02:06

3 Answers3

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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 $h(x,y-x) = f(x,y) - f(y,x)$.

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

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

Bill Dubuque
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