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

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.

I've tried very hard to solve this problem over the course of several months using the little abstract algebra I know. My first abstract algebra class starts on Monday so I hope to gain better tools to attack this problem.

My main point of attack has been to analyze various classes of algebraic equations that satisfy the condition $f(0)=0$ and $f(1)=1.$ This is because when you don't know how to think of a single object, sometimes it's better to study a class of similar objects.

Then I've reflected the entire class about $x=1/2$ to generate points in $(0,1)^2,$ and taken the collection of these points as the elements in the space. I'm specifically looking to see if I can generate points in this way, that satisfy closure w.r.t. the binary operation:$(\times).$ For example, if I have points that have algebraic numbers as coordinate pairs $P_{\in(0,1)^2}(\Bbb A,\Bbb A)$ then possibly by multiplying two points component wise, closure could be accomplished.

I don't know if this method alone will lead to a definitive answer as to whether $z$ is algebraic but it might give a helpful perspective on the road to a solution.

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    $\begingroup$ @N.S. I stated that $x\in\Bbb Q~\cap(0,1).$ $\endgroup$ – Ultradark Aug 22 at 22:08
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    $\begingroup$ A polynomial over $\mathbb Q$ you mean? $\endgroup$ – Zarrax Aug 22 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 at 22:32
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    $\begingroup$ Some context $\endgroup$ – K B Dave Aug 22 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 at 3:14

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