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Is $z=e^{\frac{1}{\log(x)}}$ a solution to a polynomial? $x\in\Bbb Q~\setminus\{0,1\}.$

Wolfram is telling me $z$ is algebraic but I'm not sure that I believe this.

I believe it would follow from Schanuel's conjecture,

$\mathbb{Q}(\ln x,\mathrm{e}^{1/\ln x})$

has transcendence degree $2$ whenever $x$ is algebraic.

Edit 9/22/2021:

Transform this into $\log(z)\log(x)=1.$ We can inspect that this is of the form of a hyperbola (if we use a change of coordinates to linearize the problem).

My intuition tells me that there are two "incompatible" structures meshed together which need to be untangled.

Let's step back and view this problem from a geometric standpoint in higher dimensional space.

I will take two distinct spaces, that each have linear structures on them. The mathematical object $\zeta^2$ is simply related to $\Bbb R^2$ in the following way:

$$ \exp: \Bbb R^2 \to \zeta^2$$ and this is a diffeomorphism (from a manifold theory standpoint alone).

Due to the "universality" of the $\exp$ map,

Postulate: Every linear structure on $\Bbb R^2$ can be re-constituted as a linear structure on $\zeta^2.$

For example $\exp$ is a group isomorphism and the group structure on $\Bbb R^2$ can be pushed onto an isomorphic group structure on $\zeta^2$. Using this key fact, a field structure can be encoded in $\zeta^2$ by way of the group structure and constructing a multiplication map. I've proved that a real vector space structure can be placed on $\zeta^2.$

So we have two disjoint objects each with linear structures on them, and a map between these objects.

What I believe is the main issue is that $z$ itself is a mixture of two incompatible linear structures (in dim. 1 as opposed to dim. 2) in a way that makes it nearly impossible to determine whether or not $z$ is algebraic.

So the objects $\Bbb R^2$ and $\zeta^2$ can be viewed as inherently different structures with their own linear structures on them and a map between the two objects.

What are we doing by inserting $\Bbb Q$ into the variable slot for $x$?

$$z=e^{\frac{1}{\log(x)}}$$

Maybe we're forcing this inherently nonlinear structure (i.e. nonlinear analytic curve with respect to $\Bbb R^2$) of $\varphi(x)=e^{\frac{1}{\log(x)}}$ to obey the classical Riemannian metric $ds^2=dx^2+dy^2.$

Using this train of ideas, I have reason to believe that thinking strictly in terms of $\Bbb R^2$ and $\zeta^2$ as different objects, that $\varphi(x)$ is an algebraic curve with respect to $\zeta^2.$

Once I embed this algebraic curve into $\Bbb R^2$ with $\Bbb R^2$'s usual metric something is lost and I can't reconcile it.

It should be clear that $\varphi(x)$ is not a polynomial in $\zeta^2$ rather its a hyperbola, so it's an algebraic curve in $\zeta^2.$

And we're inserting rational elements with respect to $\Bbb R^2$ into this algebraic curve's independent variable slot (with respect to $\zeta^2$).

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  • 1
    $\begingroup$ @N.S. I stated that $x\in\Bbb Q~\cap(0,1).$ $\endgroup$
    – geocalc33
    Aug 22, 2019 at 22:08
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    $\begingroup$ A polynomial over $\mathbb Q$ you mean? $\endgroup$
    – Zarrax
    Aug 22, 2019 at 22:31
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    $\begingroup$ I'm amazed by the close vote. I think it's pretty clear that the question is whether $z$ is algebraic. $\endgroup$
    – Rob Arthan
    Aug 22, 2019 at 22:32
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    $\begingroup$ Some context $\endgroup$
    – K B Dave
    Aug 22, 2019 at 23:52
  • $\begingroup$ Also FWIW, I think it's unlikely that we can do better than the claim $$\mathrm{e}^{u}\in\mathbb{A}\Rightarrow \{\mathrm{e}^{u^2},\mathrm{e}^{1/u^2},\mathrm{e}^{1/u}\}\not\subset\mathbb{A}\text{,}$$ as follows from the five exponentials theorem. $\endgroup$
    – K B Dave
    Aug 23, 2019 at 3:14

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