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.