Lemma 9.14.2. Let p be a prime number. Let F be a field of characteristic p. Let t \in F be an element which does not have a pth root in F. Then the polynomial x^ p - t is irreducible over F.
Proof. To see this, suppose that we have a factorization x^ p - t = f g. Taking derivatives we get f' g + f g' = 0. Note that neither f' = 0 nor g' = 0 as the degrees of f and g are smaller than p. Moreover, \deg (f') < \deg (f) and \deg (g') < \deg (g). We conclude that f and g have a factor in common. Thus if x^ p - t is reducible, then it is of the form x^ p - t = c f^ n for some irreducible f, c \in F^*, and n > 1. Since p is a prime number this implies n = p and f linear, which would imply x^ p - t has a root in F. Contradiction. \square
Comments (0)
There are also: