# proof of transcendental root theorem

###### Proposition 1.

Let $F\subset K$ be a field extension with $K$ an algebraically closed field. Let $x\in K$ be transcendental over $F$. Then for any natural number $n\geq 1$, the element $x^{1/n}\in K$ is also transcendental over $F$.

###### Proof.

Suppose $x$ is transcendental over a field $F$, and assume for a contradiction that $x^{1/n}$ is algebraic over $F$. Thus, there is a polynomial $P(y)\in F[y]$ such that $P(x^{1/n})=0$ (note that the polynomial $y^{n}-x$ is not a polynomial with coefficients in $F$, so $P(y)$ might be more involved). Then the field $H=F(x^{1/n})\subseteq K$ is a finite algebraic extension of $F$, and every element of $H$ is algebraic over $K$. However $x\in H$, so $x$ is algebraic over $F$ which is a contradiction. ∎

Title proof of transcendental root theorem ProofOfTranscendentalRootTheorem 2013-03-22 14:11:41 2013-03-22 14:11:41 alozano (2414) alozano (2414) 6 alozano (2414) Proof msc 11R04 AlgebraicElement AlgebraicClosure Algebraic AlgebraicExtension AFiniteExtensionOfFieldsIsAnAlgebraicExtension