proof of transcendental root theoremProofOfTranscendentalRootTheorem2004-02-25 17:55:162004-02-25 18:21:10Prooftranscendental root theorem0% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{prop}
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 $F$.
\end{prop}
\begin{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 {\it 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.
\end{proof}