# Is the set $\{X\sqrt{Y}, \sqrt{X}Y\}$ algebraically independent over $\mathbb{C}$?

Suppose $X$ and $Y$ are transcendental over some field $F$ (doesn't actually have to be $\mathbb{C}$, but I chose that for definiteness; I believe the answer will not depend on the exact field, as long as both $X$ and $Y$ are transcendental over $F$).

How would one go about proving that the set $\{X\sqrt{Y}, \sqrt{X}Y\}$ is algebraically independent over $F$, or finding an explicit algebraic dependence relation?

This is a special case of a more general question I have about whether having $\{\alpha, \beta\}$ and $\{\gamma, \delta\}$ algebraically dependent over $F$ implies that $\{\alpha\gamma, \beta\delta\}$ is also algebraically dependent over $F$.

Just choose $\alpha = X$, $\beta = \sqrt{X}$, $\gamma = \sqrt{Y}$, $\delta = Y$. (Note that choosing the reverse order for $\gamma, \delta$ will give the set $\{XY, \sqrt{XY}\}$ which is clearly algebraically dependent).

This isn't homework, but the question came to my mind and I tried finding an algebraic dependence relation without any success so far. Mostly my attempts consisted of squaring the expressions (getting $X^2Y, XY^2$) and trying to isolate $X$ and $Y$ in terms of these expressions.

Suppose $a=X\sqrt{Y}$ and $b=\sqrt{X}Y$ satisfy a polynomial $p(a,b)=0$
Then lets map $X\mapsto u^2$ and $Y\mapsto v^2$.
Then we get that $p(u^2v,uv^2)=0$
Let $p_n$ be the homogeneous part of $p$ of degree $n$. Then $p_n(u^2v,uv^2)=u^nv^np_n(u,v)=0$. Therefore $p_n=0$ (for all $n$).
Hence $p=0$.
• Why is $u^nv^np_n(u, v)=0$? Commented Jul 27, 2017 at 4:03