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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!

nmasanta
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furfur
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    Polynomials over what type of coefficient ring? It's not true in general. Is it a problem from calculus or abstract algebra? – Bill Dubuque Jul 22 '19 at 13:23

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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.

Suzet
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  • Doesn't $f$ have to be monic over $k[y]$ as well for polynomial division to work? – freakish Jul 22 '19 at 13:05
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    Counterexample: $1$ is antisymmetric over $\Bbb Z/2$ but is not divisible by $,x-y\ \ $ The result you use is called the Factor Theorem. Better to invoke it by name rather than reprove it inline. – Bill Dubuque Jul 22 '19 at 13:07
  • @freakish No, we only need the leading coefficient of the non-zero polynomial we are dividing by to be a unit. It may be shown by induction on the degree of the polynomial you wish to divide. – Suzet Jul 22 '19 at 13:12
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    I don't understand half of what is said here, but I think the idea is that $x^n = x^{n-1}(x-y)+yx^{n-1}$. We can repeat the process. – Jakobian Jul 22 '19 at 13:14
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    @BillDubuque You are right. If I am not mistaken, it is due to the fact that antisymmetric polynomials are not alternating polynomial in characteristic $2$, whence I can't say $f(y,y)=0$ anymore. I shall edit my answer to point this out. – Suzet Jul 22 '19 at 13:15
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    Note also for rings we have $n$ is antisymmetric in $,\Bbb Z/(2n),\ $ which is possible since the OP doesn't restrict the coef ring to fields. – Bill Dubuque Jul 22 '19 at 13:23
  • @BillDubuque You are absolutely right. Thank you for pointing this out. – Suzet Jul 22 '19 at 13:25
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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 $x^iy^j - y^ix^j = x^jy^j(x^k-y^k)$.

Magma
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    When you are working in characteristic $2$, the diagonal coefficient $c_{i,i}$ may not vanish. The polynomial $f$ could still have terms of the form $ax^iy^i$. See the conter example of Bill Dubuque in the comment section below my answer. – Suzet Jul 22 '19 at 13:40
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    Good point. More generally, $cxy$ would be a counterexample (to my solution, but also OP's statement) in any ring that has some $c$ with $c+c=0$. – Magma Jul 22 '19 at 13:49
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    (And $c \neq 0$ of course.) – Magma Jul 22 '19 at 14:12
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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: $x-y$.

Angel Moreno
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  • Your point $1$ is not true in characteristic $2$. See the conter example of Bill Dubuque in the comment section below my answer. – Suzet Jul 22 '19 at 14:04