Let G be the galois group of F over K, and let L be an intermediate extension of K within F, and let H be a subgroup of G. The extension H′ is the field fixed by the automorphisms in H. This is called the fixed field of H. The subgroup L′ is the subgroup of automorphisms fixing L.
By definition, the extension F/K is galois if G fixes exactly K. This is equivalent to the condition that some automorphism in G moves u for every u not in K. Using our notation, G′ = K.
As shown earlier, adjoining the real cube root of 2 is not galois. The only automorphism is trivial, and that fixes all of F, rather than K.
For the rest of this page, let F/K contain intermediate extensions L and M such that M contains L. Also, let H and J be subgroups of G such that J contains H. Note that subgroups and subfields need not be proper. Let 1 be the identity element in G, the trivial automorphism on F. The following relationships are not hard to verify.
An intermediate extension or subgroup x is closed if x′′ = x. Note that F is galois over L iff L is closed. Closed extensions and subgroups correspond 1-1 via the map x → x′. Apply the bidirectional map again to get x′′, which is x.
The following assumes the index of H in J is finite, or the dimension of M over L is finite.
the index of M′ in L′ ≤ the dimension of M over L. Let n be the dimension of M over L and proceed by induction on n, with n = 1 implying M = L and M′ = L′. Choose u in M-L, and let p(x) over L have degree d and root u. The dimension of L(u) over L is d and the dimension of M over L(u) is n/d. If d < n, induction finishes the proof. Otherwise d = n and M = L(u).
Let w be an automorphism in G fixing L, and let w*M′ be a left coset of M′ in L′. Any automorphism in this coset maps u to u via something in M′, then maps u to some other root of p(x) via w in L′. The action on u determines the automorphism of M over L, so the map from left cosets of M′ in L′ into the L automorphisms of M is 1-1. Thus the index of M′ in L′ ≤ n.
A completely analogous result holds for subgroups. Suppose the index of H in J is n, yet the dimension of H′ over J′ exceeds n.
Let u1 through un+1 be linearly independent elements in H′, where H′ is a vector space over J′. Let t1 through tn represent the left cosets of H in J. Apply t1 to the n+1 independent elements u1 through un+1. Write this down as a seequence of coefficients, ci = t1(ui). Then write the following homogeneous linear equation.
c1x1 + c2x2 + c3x3 + … + cn+1xn+1 = 0
The variables x1 x2 etc are unknowns, taken from the field F.
Repeat the above procedure for t2, t3, and so on. This gives a system of n linear equations in n+1 unknowns. Combine the coefficients into an n by n+1 matrix c, where ci,j = ti(uj).
Let's see - n equations and n+1 unknowns. There are indeed solutions to this system of simultaneous equations. Let the vector x be a nontrivial solution with the minimum number of nonzero entries. Permute the components of x and the columns of c, so that x1 through xr are nonzero. Scale the vector so that x1 = 1.
Let t1 be the automorphism that represents H, hence t1 fixes the independent elements in u. Since x is a solution, the sum of xiui = 0 via the first row of c. If the solution vector is all zeros, except for x1 = 1, then x1u1 = u1 = 0, which is a contradiction. There are at least two nonzero entries in the solution vector x.
If each xi lies in J′, the first r elements of u are not independent. The linear combination, with coefficients from x, produces 0. So assume x2 is not in J′. Since x2 is not fixed by J, there is some automorphism w in J that moves x2.
Write the matrix equation c*x = 0, then apply w, giving w(c)*w(x) = 0. We will see that c and w(c) are the same equations in disguise.
Apply w to t1(ui), which is the same as applying w*t1 to ui. Now w*t1 is a representative of some other coset of H in J. Perhaps w*t1 is in the same coset as t2. Applying w*t1 to ui is just like applying t2 to ui. The first row of w(c) looks just like the second row of c. Since w permutes the cosets of H in J, w(c) looks just like c, with its rows permuted. The solutions to the system of equations represented by w(c) are exactly the same solutions that satisfy c.
This means x satisfies w(c), and more important, w(x) satisfies c. Subtract, and x-w(x) satisfies c. This difference yields x1 = 0, while x2 is nonzero. This contradicts the selection of x, as having the fewest nonzero entries. Therefore the dimension of H′ over J′ ≤ n, which was the index of H in J.
Let H have finite index in J, and let H be closed. We know that H′′ = H and J′′ contains J. Thus the index of H in J ≤ the index of H in J′′ ≤ the index of H′′ in J′′ ≤ the dimension of H′ over J′ ≤ the index of H in J. Everything reverts to equality, J = J′′, and J is closed. The dimension and index are equal.
A similar argument shows that if L is closed and M is finite over L then M is closed.
If F is a finite galois extension of K, K is closed, and every intermediate extension L over K is finite, and closed.
Remember that dimension and index are equal, between closed subfields and subgroups. The dimension of F over K is the index of 1 in G, or the order of G. Therefore G is finite, as is every subgroup of G. Since 1 in G is closed, every subgroup of G is closed. All subfields and subgroups are closed, and correspond to each other. Partially order the intermediate field extensions by set inclusion, and partially order the subgroups of G by set containment. The two lattices are isomorphic, with x → x′ implementing the correspondence. pairs of comparable fields and groups have the same dimension/index.
If F is not galois over K it is certainly galois over K′′, and the dimension of F exceeds the order of its galois group. Thus F/K is galois iff the order of G = the dimension of F over K.
Let F/K be galois and let E be an intermediate extension. By correspondence, the dimension of F over E is the index of 1 in E′, which is the size of the group of automorphisms on F that fix E. Therefore F is galois over E, with galois group E′. We cannot claim E is galois over K, unless E′ is normal in G, But that's coming up next.