Inverse Galois Problem

\mathbf{Problem:} Use the fact that any finite group G is isomorphic to a subgroup of S_n to prove that given any finite group G there exists fields F and E/F such that

\displaystyle \text{Gal }E/F\cong G

\mathbf{Proof:} Any field K, x_1,x_2,\cdots,x_n are indeterminate. Let E=K(x_1,x_2,\cdots,x_n). D is the field of symmetric rational functions in latex x_1,x_2,\cdots,x_n Or if we let p_1,p_2,\cdots,p_n are basic symmetric polynomials, namely

\displaystyle p_1=\sum x_i, \displaystyle p_2=\sum _{i>j} x_ix_j, \cdots \displaystyle p_n=x_1x_2\cdots x_n,

then D=K(p_1,p_2,\cdots,p_n).

As we all know, \text{Gal }E/D=S_n. Since G\leq S_n, by the Galois corresponding theorem, there exists an intermediate field F such that \text{Gal }E/F\cong G.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} This is the easiest inverse Galois problem since one does not specify the ground field. If you restrict F=\mathbb{Q}, things are much trickier.

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