transcendental root theorem

Suppose a constant $x$ is transcendental over some field $F$. Then $\sqrt[n]{x}$ is also transcendental over $F$ for any $n\ge 1$.

Proof.

Let $\overline{F}$ denote an algebraic closure of $F$. Assume for the sake of contradiction that $\sqrt[n]{x}\in \overline{F}$. Then since algebraic numbers are closed under multiplication (and thus exponentiation by positive integers), we have ${(\sqrt[n]{x})}^n=x\in \overline{F}$, so that $x$ is algebraic over $F$, creating a contradiction. ∎

Title	transcendental root theorem
Canonical name	TranscendentalRootTheorem
Date of creation	2013-03-22 14:04:23
Last modified on	2013-03-22 14:04:23
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	8
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 11R04