Conversely, let F be a splitting field for a set of polynomials S, hence F is algebraic and normal. We may assume all polynomials in S are monic and irreducible. Further assume the polynomials in S are separable, and show F is galois as follows.
First let F be finite over K. Let G be the galois group of F over K, hence F is galois over K′′, G is finite, and |G| = the dimension of F over K′′. If |G| also equals the dimension of F over K, then K = K′′, and F is galois over K. Use induction on the dimension of F/K to prove equality, with F = K having dimension 1 and group 1.
Pick a polynomial p in S with degree d > 1, and let u be a root of p(x) in F. Let H be the subgroup of G that fixes K(u). For notational convenience, let H fix E, where E may be larger than K(u). If w represents a left coset of H in G, then w, and every other automorphism in the coset of w, moves u to another root of p(x), and that defines the action of w inside E. If different automorphisms both move u to v then apply the first and the inverse of the second and find an automorphism that fixes u, and belongs to H. So different cosets become different isomorphisms of K(u). Cosets of H in G become distinct autommorphisms of E over K.
How many ways can we move u to v? There is an isomorphism from K(u) onto K(v) for every other root v in F, and since F is normal, the isomorphism extends to all of F. Thus an automorphism in G moves u onto v for every root v in F. Since p is separable there are d distinct roots, and since F is normal they are all in F. There are d cosets of H in G, and the index of H in G agrees with the dimension of K(u) over K. By induction this holds for F/K, of dimension n.
Now let F be infinite. Select any u in F, remembering that u is algebraic. Find the polynomial associated with u and embed K(u) in the splitting field for this polynomial, which we will call E. From the above, E/K is galois. Some automorphism of E moves u to v. Extend this to an automorphism on all of F. The arbitrary element u moves, and the only thing that remains fixed is K. Therefore F is galois over K.
In summary, the extension F/K is the splitting field for a set of separable polynomials iff F is galois and algebraic.
As a corollary, any splitting field with base field K is galois, provided the characteristic of K is 0, or K is finite. All polynomials are separable, and the conditions outlined above are satisfied.
An infinite algebraic extension is galois iff it splits a set of separable polynomials, iff it is the union of finite galois extensions.