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Are there two algebraically independent irrational numbers $\alpha,\beta$ s.t. $\alpha^\beta$ is a rational number?

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  • $\begingroup$ What does “transcendentally independent” mean? $\endgroup$
    – Emil Jeřábek
    Commented May 30 at 14:02
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    $\begingroup$ I suppose you mean nonzero polynomial. Then this is normally called “algebraically independent”. $\endgroup$
    – Emil Jeřábek
    Commented May 30 at 14:06
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    $\begingroup$ I also suppose you want this to be unconditionally provable, as it is trivial to construct such examples using Schanuel’s conjecture (e.g., $\alpha=e$ and $\beta=\log2$). $\endgroup$
    – Emil Jeřábek
    Commented May 30 at 14:21
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    $\begingroup$ It's true unconditionally. An algebraic curve $P(x,y)=0$ and the analytic curve $(e^t,\frac{ln 2}{t})$ with $t>0$ has only countably many intersections. So there is a value of $t$ that is not on any algebraic curve $P(x,y)=0$ with $P$ in $\mathbb{Z}[X,Y]$. Then $\alpha=e^t$ and $\beta=\frac{ln 2}{t}$ are algebraically independent with $\alpha^\beta=2$. $\endgroup$
    – Oleg Eroshkin
    Commented May 30 at 14:34
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    $\begingroup$ @GeraldEdgar One of Gel'fond's results implies that at least one of $2^{\ln 2}$, $2^{\ln^2 2}$, $2^{\ln^3 2}$ is algebraically independent with $\ln 2$. So you can take $\alpha=2^{\ln^k 2}$ and $\beta=\ln^{-k} 2$ for one of $k=1,2,3$. I don't know if that is sufficiently explicit for you. $\endgroup$
    – Oleg Eroshkin
    Commented May 31 at 13:27

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