Question

Use mathematical induction to prove that If p(x) in F[x] and deg p(x) = n, show that the splitting field for p(x) over F has degree at most n!.

Answer 1

Proof:

We prove this result by induction on the degree of p(x). The case of n = 1 is clear as there is no extension so the degree is clearly 1. Assume inductively that for a polynomial g(x) ? E[x] where E is some field that the splitting field Eg of g(x) has degree less then or equal to (deg g(x))! whenever deg(g(x)) ? k-1. Note here that it is important to assume the result for all fields and all degrees of g(x) less then or equal to k ? 1. We will see why this is important in a moment.

Let p(x) ? F[x] be a polynomial of degree k. Let ? be a root of p(x) and consider F [?] over F . The degree of this extension is at most k and is equal to k precisely when p(x) is irreducible. We can factor p(x) = (x ? ?)^m g(x) in F [?][x]. The degree of g(x) is k ? m and so we can apply our induction hypothesis to g(x) and the field E = F[?]. So the splitting field Eg of g(x) is a finite extension of F [?] of degree at most (k ? m)!. Since Eg contains ? and all the roots of g(x), it must contain all the roots of p(x). In particular, the splitting field of p(x)must be contained in Eg. However, we see that we have F= ?F[?]?Eg and we have[Eg :F]=[Eg :F[?]][F[?]:F]? k(k ? m)! ? k(k ? 1)! = k!. Now using that the splitting field of p(x) is a subfield of Eg, it must be its degree over F is less then or equal to the degree of Eg over F, i.e., the degree of the splitting field of p(x) over F is at most k!. Thus we are done by induction. ??