How do I prove an antisymetrical polynomial $~f(x,y)~$ is divisible by $~x-y~$

Question

An antisymetrical polynomial $~f(x,y)~$ is defined such that $f(x,y)=-f(y,x).$ How do I prove there exists a polynomial $~g(x,y)~$ such that $f(x,y)=g(x,y) \cdot (x-y)$

I tried by letting $f(x,y)=x^n \cdot g_n(x,y)+...+g_0(x,y)$ but I didn’t know what to do after.

Please help me!

Answer 1

Edit: the following works whenever the base ring is a field of characteristic not $2$.

One may look $f$ as a univariate polynomial is the ring $k[y][x]$ just the way you did. This is a ring of polynomials not over a field, but over a domain. Such a polynomial ring is not euclidian in general, but it still has a notion of euclidian division which works in particular whenever we divide by a polynomial whose leading coefficient is a unit.

This is the case of the monic degree $1$ polynomial $x-y$. Hence, arguing this way, we know that there exists polynomials $g\in k[y][x]$ and $r\in k[y][x]$ such that $$f = g(x-y) + r$$ along with the condition $\deg_x(r)<\deg_x(x-y)=1$. Hence, $r=r(y)$ is in fact independant on $x$. Now, let us set $x=y$ in our equation. Because $f$ is antisymmetric, we know that $f(y,y)=0$ (assuming that the characteristic of $k$ is not $2$). Thus, we obtain $$r(y)=0$$ which proves that the univariate polynomial $r$ is in fact $0$. This gives the conclusion we wanted.

Answer 2

Elementary solution: Let

$$f(x,y) = \sum_{i,j}c_{ij}x^iy^j.$$

Since $f$ is antisymmetric, we find $c_{ij} = -c_{ji}$, so we can write $$f(x,y) = \sum_{i>j}c_{ij}(x^iy^j-y^ix^j).$$

Now we've found out that $f$ is some linear combination of terms $x^iy^j-y^ix^j$. If we show that each of these terms is divisible by $x-y$, then so is $f$, and we're done.

So suppose $i > j$, and let $k = i - j$. Since

$$x^k - y^k = (x-y)(x^{k-1} + x^{k-2}y + x^{k-3}y^2 + \ldots + y^{k-1})$$

is divisible by $x-y$, so is $x^iy^j - y^ix^j = x^jy^j(x^k-y^k)$.

Answer 3

The polynomial $f$, by antisymmetric, can be deduced two things:

1.- The monomials of the type $Ax^ay^a$ do not belong to "f" (its zero coefficient)

2.- If a monomial of the type $Ax ^ ay ^ b, a\neq b $ belongs to f then the antisymmetric monomial of that $-Ax ^ by ^ a$ also belongs to f.

All pairs of antisymmetric monomials $Ax ^ a y ^ b -Ax ^ b y ^ a$ are divisible by $(x-y)$ very easily:

Let $c = min (a, b)$

$$\frac{Ax ^ a y^ b -Ax ^ by ^ a} {x-y} =$$ $$A x ^ c y ^ c \frac{x ^ {a-c} - y ^ {a-c}}{x-y} $$

But: $$\frac {x ^ n-y ^ n} {x-y} = x ^ {n-1} -x ^ {n-2}y + x ^ {n-3}y ^ 2 ... +(-1)^{n-1} y ^ {n-1}$$

Therefore, all pairs of antisymmetric monomials are divisible by: $x-y$.

Which implies that every antisymmetric polynomial is divisible by: $x-y$.