Prove that every finite group occurs as the Galois group of a field extension of the form $F(x_{1}, \dots , x_{n})/F$. I've seen that every abelian group is a Galois group over $\mathbb{Q}$ for some subfield of a cyclotomic field, but I'm not sure about this more general result.
 A: Recall that the Galois group of a generic $n$th degree polynomial $p$ is $S_n$, and that any group of order $n$ occurs as a subgroup of $S_n$. Let $E=\mathbb Q(x_1,\ldots,x_n)$ where $x_i$ are the roots of $p$. By the Fundamental Theorem of Galois Theory we have some subfield $F$ such that $\mathrm{Gal}(E/F)=G$. Since $E=F(x_1,\ldots,x_n)$, this is the desired extension.
A: Let $x_1,\cdots,x_n$ be indeterminates and $e_i$ the elementary symmetric polynomials in them. The field extension $K(x_1,\cdots,x_n)/K(e_1,\cdots,e_n)$ is is the splitting field of a separable "generic polynomial" given by $(T-x_1)\cdots(T-x_n)$ (invoke Vieta's formulas to see the coefficients are the $e_i$s), hence is Galois, and so the Galois group is a subgroup of $S_n$. (Here $K$ is any field.) Observe that indeed every permutation defines an automorphism, so the Galois group is all of $S_n$.
Let $G$ be a finite group for which $G\hookrightarrow S_n$ (this works for any finite $G$ by Cayley's theorem). Take $F$ to be the fixed field $K(x_1,\cdots,x_n)^G$ so that $F(x_1,\cdots,x_n)/F$ has Galois group $G$ by the fundamental theorem of Galois theory.
