C. Solution of Equations by Radicals.

The extension field E over F is called an extension by radicals

 

 

if there exist intermediate fields where each a; is a root of an equation of the form

aj t BU1 . A polynomial f ( x ) in a field F is said to be solvable by radicals if its splitting field lies in an extension by radicals. We assume unless otherwise specified that the base field has characteristic 0 and that F contains as many roots of unity as are needed to make our sub­sequent statements valid.

 

where q takes all values in the group of automorphisms of Bj over B o, then f1 is in BQ, and adjoining successively the roots

brings us to an exten­

Let us remark first that any extension of F by radicals can always be extended to an extension of F by radicals which is normal over F. Indeed Bj is a normal extension of B0 since it contains not only ai? but fa,, where {is anv n, -root of unitv, so that B, is the splitting field

sion of B2 which is normal over F. Continuing in this way we arrive at an extension of E by radicals which will be normal over F. We now prove

THEOREM 5. The polynomial f(x) is solvable by radicals if and only if its group is solvable.

Suppose f(x) is solvable by radicals. Let E be a normal exten­sion of F by radicals containing the splitting field B of fl(x), and call G the group of E over F. Since for each i, B. is a Kummer extension of Bi l, the group of B. over BM is abelian. In the sequence of groups

preceding since is the group of E over B^j and Bi is a normal extension of s the group of and hence each is a normal subgroup of the

 is abelian. Thus G is solvable. However, GB is a normal subgroup of G, and G/G0 is the group of B over F, and is therefore the group of the polynomial f(x). But G/GB is a homomorph of the solvable group G and hence is itself solvable.

On the other hand, suppose the group G of f(x) to be solvable be a and let E be the splitting field. Let sequence with abelian factor groups. Call Bj the fixed field for Gr Since Gj j is the group of E over Bj x and G. is a normal subgroup of G. j, then is normal over BM and the group Gi l/Gi is abelian. Thus B. is a Kummer extension of B. ,, hence is splitting field of a polynomial of the form SO that by forming the successive splitting fields of the we see that Bj is an extension of B1 l by radicals, from which it follows that E is an extension by radicals.

Remark. The assumption that F contains roots of unity is not necessary in the above theorem. For if f(x) has a solvable group G, then we may adjoin to F a primitive nth root of unity, where n is, say, equal to the order of G. The group of f(x) when considered as lying in F' is, by the theorem on Natural Irrationalities, a subgroup G1 of G, and hence is solvable. Thus the splitting field over F1 of f(x) can be obtained by radicals. Conversely, if the splitting field E over F of f(x) can be obtained by radicals, then by adjoining a suitable root of unity E is extended to E' which is still normal over F1 . But E' could be obtained by adjoining first the root of unity, and then the radicals, to F; F would first be extended to F 1 and then F 1 would be extended to E ' . Calling G the group of E ' over F and G 1 the group of E 1 over F 1 . we see that G 1 is solvable and G/G 1 is the group of F 1 over F and hence abelian. Thus G is solvable. The factor group G/GE is the group of f(x) and being a homomorph of a solvable group is also solvable.