Galois Theory, field of rational functions
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From: ar0 on 17 May 2010 13:24 Hi, let F be a field and L := F(X_1, ..., X_n) be the field of rational functions. Now the symmetric group S_n naturally induces F-linear automorphisms via: forall f in S_n: f(a) = a forall a in F and f(X_i) = X_f(i) Now, if I define K to be the fixed field of S_n (the field of rational symmetric functions), then appearently (that's my question pretty much) L/K is a galois-extension of degree n! = |S_n| and Gal(L/K) = S_n. Now, how can I prove that L/K is indeed a galois extension? In particular: how do I prove that L/K is finite? Because then it would follow from the definition that L/K is galois. Is there a canonical way to construct a K-basis for L? -- Sick nature.
From: Arturo Magidin on 17 May 2010 14:39
On May 17, 12:24 pm, no...(a)nospam.invalid (ar0) wrote: > Hi, > > let F be a field and L := F(X_1, ..., X_n) be the field of rational functions. Now the symmetric > group S_n naturally induces F-linear automorphisms via: > forall f in S_n: f(a) = a forall a in F and f(X_i) = X_f(i) > Now, if I define K to be the fixed field of S_n (the field of rational symmetric functions), > then appearently (that's my question pretty much) L/K is a galois-extension of degree > n! = |S_n| and Gal(L/K) = S_n. > > Now, how can I prove that L/K is indeed a galois extension? > In particular: how do I prove that L/K is finite? Because then it would follow > from the definition that L/K is galois. This is a consequence of Artin's Theorem. Artin's Theorem: Let F be a field, and let G be a group of automorphisms of F. Let K be the fixed field of G. Then F is a Galois extension of K. If G is finite, then F is a finite dimensional Galois extension of K with Galois group Gal(F/K) = G. Proof. Certainly G is a subgroup of Aut(F/K). Therefore, the fixed field of Aut(F/K) is contained in the fixed field of G, which is K. But the fixed field of Aut(F/K) contains K by definition, so the fixed field of Aut(F/K) is K. This proves that F is Galois over K (that is, that the fixed field of Aut(F/K) is K; if you have a different definition of "Galois extensions", please say which one you are using). Now assume that G is finite. Then we know that the degree of F over K is equal to [F:K] = [1':G'] (1' is the fixed field of the trivial subgroup, G' is the stabilizer of G). But we always have that if I and J are subgroups of Aut(F/K), with I<J, then [I':J'] <= [J:I]. Thus we have [F:K]=[1':G']<=[G:1]=|G|, so the extension is of degree at most | G|. Since G is a subgroup of Aut(F/K) and F is Galois over K, |G|<=| Aut(F/K)|=[F:K] <= |G|, giving equality across the board, and in particular G = Aut(F/K), as desired. QED. > Is there a canonical way to construct a K-basis for L? The fixed field is the field of symmetric functions. To find a basis, see the recent discussion: http://groups.google.com/group/sci.math/browse_frm/thread/75b5fb60bae34038/886331729a8e0682?#886331729a8e0682 -- Arturo Magidin |