# transcendental root theorem

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

###### 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 TranscendentalRootTheorem 2013-03-22 14:04:23 2013-03-22 14:04:23 mathcam (2727) mathcam (2727) 8 mathcam (2727) Theorem msc 11R04