transcendental root theorem

Suppose a constant $x$ is transcendental over some field $F$ . Then $\\sqrt[n]{x}$ is also transcendental over $F$ for any $n\\geq 1$ .

Proof.

Let $\\overline{F}$ denote an algebraic closure of $F$ . Assume for the sake of contradiction that $\\sqrt[n]{x}\\in\\overline{F}$ . Then since algebraic numbers are closed under multiplication (and thus exponentiation by positive integers), we have $(\\sqrt[n]{x})^{n}=x\\in\\overline{F}$ , so that $x$ is algebraic over $F$ , creating a contradiction.∎

单词	TranscendentalRootTheorem
释义	transcendental root theorem Suppose a constant $x$ is transcendental over some field $F$ . Then $\\sqrt[n]{x}$ is also transcendental over $F$ for any $n\\geq 1$ . Proof. Let $\\overline{F}$ denote an algebraic closure of $F$ . Assume for the sake of contradiction that $\\sqrt[n]{x}\\in\\overline{F}$ . Then since algebraic numbers are closed under multiplication (and thus exponentiation by positive integers), we have $(\\sqrt[n]{x})^{n}=x\\in\\overline{F}$ , so that $x$ is algebraic over $F$ , creating a contradiction.∎
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