proof of transcendental root theorem

Proposition 1.

Let $F\mathrm{\subset }K$ be a field extension with $K$ an algebraically closed field. Let $x\mathrm{\in }K$ be transcendental over $F$. Then for any natural number $n\mathrm{\ge }\mathrm{1}$, the element $x^{\mathrm{1}\mathrm{/}n}\mathrm{\in }K$ 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
Canonical name	ProofOfTranscendentalRootTheorem
Date of creation	2013-03-22 14:11:41
Last modified on	2013-03-22 14:11:41
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	6
Author	alozano (2414)
Entry type	Proof
Classification	msc 11R04
Related topic	AlgebraicElement
Related topic	AlgebraicClosure
Related topic	Algebraic
Related topic	AlgebraicExtension
Related topic	AFiniteExtensionOfFieldsIsAnAlgebraicExtension