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[parent] proof of transcendental root theorem (Proof)
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. $ \qedsymbol$



"proof of transcendental root theorem" is owned by alozano.
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See Also: algebraic, algebraic closure, algebraic, algebraic extension, a finite extension of fields is an algebraic extension


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Cross-references: algebraic extension, finite, coefficients, polynomial, algebraic, contradiction, natural number, transcendental, field, algebraically closed, field extension
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This is version 3 of proof of transcendental root theorem, born on 2004-02-25, modified 2004-02-25.
Object id is 5625, canonical name is ProofOfTranscendentalRootTheorem.
Accessed 1492 times total.

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AMS MSC: 11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
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