Let G be the galois group of F/K and let H be the subgroup that fixes E. Borrowing notation from the previous page, H = E′. Note that F/K need not be a galois extension. But it has a galois group G nonetheless, with H as subgroup. If E is stable then H is a normal subgroup of G. Apply any automorphism c from G, which keeps the elements of E inside E, then apply something from H, which leaves E alone. Finally apply c inverse, which moves the elements of E back where they started from. The result is an automorphism that fixes E, another automorphism in H, hence H is normal in G.
Conversely, assume H is normal in G and let E be the subfield fixed by H, H′ in our notation. By normality, c*H/c returns u in E to its original position. Suppose c(u) = v for some v outside of E. Some automorphism in H moves v, else v would belong to E. Apply this automorphism and move v to w, then apply c inverse. The composite automorphism does not put u back where it started, and does not fix E. Yet H is normal, so cH/c should fix E. Therefore c keeps E inside E, and E is stable.
Next assume F is galois over K and E is stable. Every u in E-K is moved by some automorphism of F, and this automorphism restricted to E is a valid automorphism of E, hence E is galois over K. A stable extension in a galois extension is galois.
Conversely, if E is algebraic, and galois over K, then E is stable within any larger extension. For any u in E-K, let p(x) be the irreducible polynomial with p(u) = 0, and let u1 u2 … be the distinct roots of p that are present in E. Let q(x) be the monic polynomial with precisely these roots, and only one copy of each root. For any automorphism c of E, c permutes the roots of q, and fixes the coefficients of q(x). This holds for all the K automorphisms of E. Only elements in K are fixed, so q is a polynomial over K, and q divides p. Since p is irreducible, q = p. All the roots of p(x) are present in E, and E is a normal extension. Furthermore, all the roots of p have multiplicity 1, just like their counterparts in q, so E is separable. (You can read all about separable and inseparable fields next, if you wish.) Finally, an isomorphism that acts on E must move u to one of its conjugates, so the image of u lies in E, and E is stable. Every algebraic galois extension is normal, separable, and stable.
Let E be stable in the algebraic galois extension F/K. Thus E′ is a normal subgroup H of G. Let Q be the quotient group G/H. Map G onto Q by restricting the corresponding automorphism down to E. Since E is stable, this gives an automorphism on E. And since the automorphisms of E are extendable, the map is onto. Therefore the galois group of E over K is Q, or G/H.
In summary, let E be an intermediate extension in the algebrai galois extension F/K, and let E′ = H. Now E is stable iff E is galois over K, iff H is a normal subgroup of G; and the galois group of E over K is G/H.