Algebraic independence of even polynomials

Question

Fix an integer $k\in\mathbb{N},k>2$. Are the two polynomials $x^{2k}+y^{2k}+z^{2k}$ and $x^{2}+y^{2}+z^{2}$ algebraically independent? More specifically, could I write the first as a polynomial function of the second?

The context of my question is that I am trying to understand what the elements in the center of the universal enveloping algebra of $\frak{su}(2)$ look like. Since the rank is $1$ there is a single generator which is $x^{2}+y^{2}+z^{2}$. I would like to know whether $x^{2k}+y^{2k}+z^{2k}$ is also in the center.

Answer 1

The answer is negative. Assume contrariwise that there would exist a polynomial $P(T)\in\Bbb{C}[T]$ such that $P(x^2+y^2+z^2)=x^{2k}+y^{2k}+z^{2k}$. Evaluating at $x=y=z=0$ we see that $P(0)=0$. However, at $x=1,y=1,z=\sqrt{-2}$ we have $x^2+y^2+z^2=0$ but $x^{2k}+y^{2k}+z^{2k}=2-(-2)^k\neq0$. This is a contradiction.

Answer 2

It suffices to show $x^{2k}+y^{2k}$ and $x^2+y^2$ are algebraically independent.

For this, we show that $x, y$ are algebraic over $\mathbb C(x^{2k}+y^{2k}, x^2+y^2)$, therefore since the transcendence degree of $\mathbb C(x,y)$ is $2$ over $\mathbb C$, we know that $\{x^{2k}+y^{2k}, x^2+y^2\}$ must be a transcendental basis of $\mathbb C[x,y]$ over $\mathbb C$.

Note that $\{x^2+y^2, xy\}$ is a transcendental basis, as $x,y$ both satisfy $T^4-(x^2+y^2)T^2+(xy)^2$. Since $x^{2k}+y^{2k}$ is symmetric of $x^2$ and $y^2$, it's a polynomial of $x^2+y^2$ and $x^2y^2=(xy)^2$, hence $x^{2k}+y^{2k}=f(x^2+y^2, xy)$. Using Jyrki Lahtonen's argument, $x^{2k}+y^{2k}$ is not a polynomial of $x^2+y^2$, hence $xy$ has a nontrivial presence in $f$, thus must be algebraic over $\mathbb C(x^{2k}+y^{2k}, x^2+y^2)$.