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Suppose $x,y,z\neq 0$, that $(x,y,z)$ is a point in $\mathbb{R}^3$ and that $td[\mathbb{Q}(x,y,z):\mathbb{Q}]=2$ (where $td[,]$ denotes the transcendence degree of the field extension).

Is it true that the set $\{x/z,y/z\}$ is algebraically independent over $\mathbb{Q}$?

Edit: I'm really interested in the case when $(x,y,z)$ is a point on the unit sphere (about the origin). In trying to make the question more simple, I've probably missed several special cases where the answer to the question as stated is obvious. One such example is given by Wojowu's comment below.

To avoid all special cases, probably what I want is to assume that $td[\mathbb{Q}(x,y):\mathbb{Q}]=2$, $td[\mathbb{Q}(x,z):\mathbb{Q}]=2$ and $td[\mathbb{Q}(y,z):\mathbb{Q}]=2$.

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  • $\begingroup$ I might be missing something, but wouldn't taking $x,y$ equal provide a counterexample to this claim whenever $x,z$ are algebraically independent? $\endgroup$
    – Wojowu
    Sep 21, 2015 at 18:07
  • $\begingroup$ You're right, I'll edit. $\endgroup$
    – user62562
    Sep 21, 2015 at 18:37
  • $\begingroup$ With all the restrictions now in place, I feel that the answer might be "yes". Any algebraic relation between $x/z$ and $y/z$ is equivalent to a homogeneous algebraic relation between $x$, $y$, and $z$ (just by multiplying by the appropriate power of $z$). But combining that relation with $x^2+y^2+z^2=1$ should force the overall transcendence degree down to $1$. I'm not learned enough to turn this intuition into a rigorous proof, though. $\endgroup$ Sep 22, 2015 at 16:35

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No: take for example $x=\alpha^2$, $y=\beta^2$, and $z=\alpha\beta$, where $\alpha$ and $\beta$ are transcendental numbers that are algebraically independent.

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    $\begingroup$ Thanks, is it known that $\{\pi^2,e^2\}$ is algebraically independent? $\endgroup$
    – user62562
    Sep 21, 2015 at 20:53
  • $\begingroup$ Hmmm, it appears that's still an open question (equivalent to $\{\pi,e\}$ being algebraically independent). But the construction is sound as an answer to your question—just replace $\pi$ and $e$ by two numbers hypothesized to be transcendental and algebraically independent. $\endgroup$ Sep 21, 2015 at 23:53
  • $\begingroup$ Thanks, if you want to edit your answer to say that then I will accept it. Btw, your construction doesn't seem to (unless I'm being silly) adapt to answer the question I'm really interested in (see edit) where the point $(x,y,z)$ satisfies $x^2+y^2+z^2=1$. Do you know if the answer is negative there as well? $\endgroup$
    – user62562
    Sep 22, 2015 at 6:49
  • $\begingroup$ @user62562 You could ask this as a new question, but I wonder why didn't do this from the beginning? $\endgroup$
    – user26857
    Sep 22, 2015 at 20:14
  • $\begingroup$ @user26857 my initial feeling was that the sphere wasn't important (and if I could see the answer for the sphere I'd also be curious about other surfaces/algebraic sets in higher dimensions) and my hope was that despite the paucity of results on algebraic independence there might be something general about 'division'. Admittedly I was wrong! $\endgroup$
    – user62562
    Sep 22, 2015 at 20:22

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