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.∎

单词	ProofOfTranscendentalRootTheorem
释义	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.∎
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