In Galois theory, some algebraic extensions are more useful than others. Typically, there are two important attributes an extension must have to satisfy the nice properties Galois theory can prove: normality, and separability. I’ll deal with normality first, and with separability in a later post.
A field extension L/K is called normal if whenever a polynomial f in K[x] is irreducible, it either has no root in L or splits into linear factors in L.
At this stage, it’s simplest to give an example of a non-normal extension: Q(2^(1/3))/Q, where the polynomial x^3 – 2 is irreducible over Q but only factors as (x – 2^1/3)(x^2 + (2^1/3)x + 4^1/3) in Q(2^(1/3)). The easiest way of seeing this is that the three roots of x^3 – 2 are 2^1/3, (2^1/3)w, and (2^1/3)w^2, where w^3 = 1; the first is real, so the field Q(2^(1/3)) is contained in R, but the other two are proper complex.
Every quadratic extension, i.e. an extension of degree 2, is normal. The proof of that is based on the fact that if an irreducible polynomial over K of degree n has a root in L, then [L:K] >= n. As a consequence of that, if f is irreducible of degree at least 3 then it has no root in L; if it is irreducible of degree 2 and has a root in L, then clearly it splits into two linear factors.
An equivalent condition for normality is that for every a in L, m(a) splits into linear factors over L. This condition is obviously necessary, since m(a) is irreducible over K; it’s also sufficient, since if f is irreducible and has a root a in L, then it must be m(a).
From there, it immediately follows that every normal extension is a splitting field. If f(i) is a family of polynomials over K, then the splitting field of f(i) over K is the smallest extension of K containing all roots of the polynomials. If the family is finite, then multiplying all its members yields one polynomial, f, such that L is the smallest field containing all roots of f.
In fact, every polynomial f over K has a unique splitting field, of degree at most n! where n = deg(f). Existence is relatively easy: given an irreducible polynomial over K, f, it’s always possible to take one root of f and adjoin it to K by letting L = K[x]/(f). Given a reducible polynomial, reduce it to irreducible factors and adjoin the root of one of them to K. After obtaining L, reduce f to irreducibles over L and continue the process until all resulting polynomials are linear.
Each step reduces the total degree of all polynomials involved by 1 and multiplies the degree of the extension of K by at most n–k+1, where k is the number of the step, so the total degree is at most n!.
For uniqueness, first note that K[x]/(f) is clearly unique. The only step that involves a choice is choosing which irreducible factor of a polynomial to extract a root out of. To prove that splitting fields are unique, it’s then enough to show that adjoining a root of f first followed by a root of g induces the same extension as adjoining a root of g first followed by a root of f.
But to see that this is the case, let L be K[x]/(f), M be L[x]/(g) (or a factor of g if g is reducible over L), L’ be K[x]/(g), and M’ be L'[x]/(f). Over M and M’, both f and g have a root, so both M and M’ contain subfields isomorphic to L and L’. Further, f has a root in the extension M/L’, so M contains M’; similarly, M’ contains M. But that implies that [M:K] = [M’:K], so that M = M’.
If L is a splitting field for f over K, then L/K is normal. To see why, let a be any element of L, and let M be the splitting field for m(a); it’s enough to prove that M = L. If [M:L] > 1, that is if m(a) has a root a‘ not in L, then K(a) = K(a‘), so that the splitting field for f over K(a), L, is equal to the splitting field for f over K(a), L’. But that’s a contradiction, since on the one hand L = L’ but on the other L’ contains L, the splitting field for f over K, as well as additional elements generated by a‘. So [L:L] = [L’:L] > 1, which is absurd.
If L/K is any field extension, then L has a smallest extension that is normal over K. That extension is defined as the splitting field of all minimal polynomials over K of elements of L. We call that the normal closure of L over K.
Finally, let M/L/K be a tower of extensions. If M/K is normal then so is M/L for fairly obvious reasons. But all other possibilities are false: it’s possible for M/K to be normal but for L/K not to be normal, e.g. if K = Q, L = Q(2^1/3), M = Q(2^1/3, w); and it’s possible for M/L and L/K to be normal but for M/K not to be, e.g. if K = Q, L = Q(2^1/2), M = Q(2^1/4).