Existence of a polynomial in $C[x,y]$

Question

Let $f ∈ C[x, y]$ be such that $f(x, y) = f(y, x)$. We have to show that there is a $g ∈ C[x, y]$ such that $f(x, y) = g(x + y, xy)$.

I was trying to define a function $g(x,y)$ but unfortunately failed.

Any hint will be helpful.

Answer 1

Apply induction to the degree of $f$ and to the number of terms of $f$ of largest degree. What you want to show is that for some polynomial $g_0$, the polynomial $f(x,y) - g_0(x+y,xy)$ has (at least) one less term with the largest degree and so by induction $f(x,y) - g_0(x+y,xy) = g_1(x+y,xy)$ and you take $g = g_0 + g_1$.

So suppose that $cx^ay^b$ is a term of $f$ with largest degree. Now because $f$ is symmetric, $cx^by^a$ is also a monomial. So without loss of generality we can assume that $a \le b$. Now let us factor out $(xy)^{a}$ to obtain

$$cx^ay^b + cx^by^a = c(xy)^a( y^{b - a} + x^{b - a} ). \tag{1} $$

Also $$x^n + y^n = (x + y)^n - \sum_{i = 1}^{n - 1} \binom{n}{i} x^iy^{n-i} = (x + y)^n - (xy)\sum_{i = 1}^{n - 1} \binom{n}{i} x^{i-1}y^{n-i-1}. \tag{2} $$

So combining $(1)$ and $(2)$ you can remove a polynomial in $xy$ and $x + y$ from $f$ so that $f(x,y) - g_0(x+y,xy)$ has total degree no more than $\deg f$ and $f(x,y) - g_0(x+y,xy)$ has at least one less term of largest degree.