Exercise 17, § 14

Let K be a field, $\Omega$ an algebraically closed extension of K and x an element of $\Omega$ transcendental over K. Show that for every prime number q there exists a subextension M of $\Omega$ such that $M(x)$ is a Galois extension of degree q of $M$. (Consider the element $y = x^q$ if q is not the characteristic of K, $y = x^q - x$ in the opposite case, and a subextension L of $\Omega$ generated by a transcendence basis of $\Omega$ over K containing y and besides the q-th roots of unity if $y = x^q$; use Ex. 12 of V, p. 163.)