Harder than I thought. A beginning:
Sketch of Galois theory
Galois theory translates a problem about solutions of polynomials over a field, into a problem in group theory, and the analysis uses the theory of dimension of abstract vector spaces, quotients of polynomial rings, and quotients of groups. Hence one must know something of all those subjects to even understand the statements.
Groups and fields
A (concrete) group G s a collection of permutations of the elements of some set S, together with their inverses, and which is closed under composition; an abstract group is a set G with an associative operation with an identity s.t. all elements have inverses. [An abstract group can be viewed concretely as a group of permutations of its own elements via the given operation.]. G is called abelian if the operation is commutative.
E.g.the isometries of a square form a non abelian group of order 8, [a dihedral group]; the full collection of permutations of a set of n elements is a non abelian group of order n! called Sn = the symmetric group on n elements; the rotation group of a cube is the group S4 of order 24; the integers Z are an (infinite) abelian group under addition, the rational numbers Q are a group under addition, and the non zero rationals are an abelian group under multiplication. The real numbers and the complex numbers are also abelian groups under addition whose non zero elements are abelian groups under multiplication. The complex nth roots of 1, i.e. the n complex solutions of X^n-1, form an abelian multiplicative group of order n, consisting of the powers of any “primitive” root, and isomorphic to the additive group Z/nZ. Any number system F, with two operations, +, . , which is an abelian group under addition, and its non zero elements are an abelian group under multiplication, (s.t. multiplication distributes over addition), is called a field. We assume all our coefficient fields contain Q.
Splitting fields of polynomials
If F is any field, and f is a polynomial with coefficients in F, a root of f is an element ß of a field K containing F, and such that in K, f(ß) = 0. If f has degree n, and K contains F and n roots of f, and no smaller subfield of K contains F and all n roots, then K is called a root field, or solution field, or splitting field, of f over F. E.g. the complex number field C is the splitting field over the real field R, of the polynomial X^2+1. Hence the complex field can be viewed as the quotient R[X]/(X^2+1) = {real polynomials in X, but equated if they become equal when X^2 is set equal to -1} = { expressions of form a+bi with a,b real and i^2 = -1}.
Similarly, the splitting field over Q of X^2-2 is the set K = {a+bs, with a,b, rational and s^2 = 2, i.e. with s = sqrt(2)}. Then K ≈ Q[X]/(X^2-2) = {rational polynomials in X, but equated if they become equal when X^2 is set = 2}. These “quadratic” splitting fields K have vector dimension over their base field F, equal to 2, the degree of the irreducible polynomial over F from which K is obtained by adjoining one root of f. In general the field F[X]/(f) obtained by adjoining one root of the irreducible polynomial f, has vector dimension over F equal to the degree n of f, and the full root field K obtained by adjoining all n roots of f, has vector dimension d over F, where n divides d and d divides n!.
I.e. to obtain the full root field of f over F, one first adjoins one root, getting a field isomorphic to the quotient F[X]/(f) = F1 = F(ß), where ß = X(modf). Then in this extension field, the polynomial f has at least one root, so in F1, the polynomial f has at least one linear factor. Unfortunately, since in this quotient field, the letter X has been set equal to that root, we need a new letter for the variable of f thought of again as a polynomial. I.e. in F1, f(X(modf)) = (f(X))(modf) = 0. So let we could let f(Y) now be the polynomial with coefficients in F1, the same coefficients as f had, but now in F1. But that seems silly, so instead of thinking of our field F1 as the actual quotient F[X]/(f(X)), just pick a new letter ß1 for the root, and write the field F1 as F(ß1). Then think of f(X) as a polynomial in the variable X, with coefficients in this F1 = F(ß1). E.g. if F = Q, just pick ß1 as some complex root of f.
Anyway we have f(X) = (X-ß1)g(X) with g a polynomial with coefficients in F1. Now the question is whether g is irreducible over F1, or whether it factors in some way. If we are really lucky, maybe g already factors completely into linear factors over F1, and then F1 is already the full splitting field of f over F. Then the splitting field has vector dimension over F equal to n = degree(f). At the opposite extreme, if g is irreducible, and after adjoining a root ß2 of g, we get a field F2 where g factors as g = (X-ß2)h with h still irreducible, and suppose this keeps up, i.e. suppose every time we adjoin another root, we only get one linear factor and the other factor is always still irreducible. Then it takes n-1 steps to get a full splitting field of f, and the vector dimension of that splitting field over f is n!, where n= degree(f).
E.g. one extreme case happens for the polynomial f(X) = X^4+X^3+X^2+X+1, over Q, where (X-1)f(X) = X^5-1. If we adjoin one root ß of f, we get the field F(ß) = F1 of dimension 4 over Q, and we claim this field already contains all 4 roots of f. That is easy, since ß is a primitive 5th root of 1, (because 5 is prime), so its powers ß, ß^2, ß^3, ß^4, ß^5 = 1, are all 5 roots of X^5-1, and in particular the first 4 of those give all 4 roots of f. Thus the splitting field of f over Q has minimum vector dimension over Q, i.e. equal to n = degree(f).
To get an example where the splitting field has vector dimension n! over Q, where degree(f) = n, we can take f(X) = X^3-2. Then adjoining one root gives us a field isomorphic to the subfield of R generated by Q and the one real cube root ß of 2. I.e. all fields obtained by adjoining one root of f, are isomorphic to the quotient field Q[X}/(X^3-2), so we can take any one of them as a model. Thus we can regard F1 = F(ß) as a subfield of the reals, which makes it obvious that F(ß) does not contain any more roots of f, since those are not real. Hence to get the full splitting field of f over Q, we need to adjoin another (hence both) root of a quadratic polynomial over F1. This full splitting field F2 hence has dimension 2 over F1, so the dimension of F2 over Q is 6 = 3!, where degree(f) = 3.
For an intermediate example, take f(X) = X^4-2, over Q. Then adjoining a real 4th root ß of 2, gives also -ß as a second root, and thus f factors over Q(ß) into (X-ß)(X+ß).g(X) = (X^2-ß^2)(X^2+ß^2), where g = (X^2+ß^2) = (X^2 + sqrt(2)), is an irreducible quadratic with 2 imaginary roots. Thus the full splitting field of f over Q has vector dimension 8; i.e degree(f) = 4, so we have 4 < 8 < 4!.
The basic problem of solvability of equations by radicals
If f(X) is a polynomial with coefficients in a field F, we ask whether one can write a root the roots of f as an algebraic combination of elements of F and “radicals”. In the language of fields as given above, this asks whether a splitting field K for F, is contained in a field obtained from F by successively adjoining roots of equations of form X^r - c, where c is an element of a previously obtained field. I.e. a field extension is called radical if the larger field is obtained from the smaller field as the splitting field of a polynomial of form X^r - c, where c is an element of the smaller field, i.e. by adjoining rth roots of an element c. Then the question is whether K contained in a union of radical extensions, starting from F. Galois theory implies that if we can choose f irreducible over F, and so that the vector dimension of K over Q is n!, where n = degree f, then the answer is no for n ≥ 5. His proof depends on analyzing the “Galois group” of symmetries of K over F. Briefly, a union of radical extensions has a symmetry group that decomposes as a union of subgroups with abelian quotients, but this is not always the case for a general splitting extension.
More specifically, if n = degree(f), the Galois group G is a subgroup of Sn, with order(G) equal to the vector dimension over F of the splitting field K of f. Hence if that dimension equals n!, then G ≈ Sn, which does not have such an abelian quotient decomposition for n ≥ 5.
