Let F = K(u), where up = a, for some a in K. Thus u is a root of xp-a, a polynomial that is, by assumption, irreducible. Note, sometimes K is the fraction field of a ufd, and a is a nonunit in that ufd, divisible by some prime factor q, but not q2, whence the polynomial becomes irreducible by eisenstein's criterion.
Assume there is an element r in K that is a primitive pth root of 1. Thus K contains all the pth roots of 1, and splits xp-1.
The elements u, ur, ur2, etc, are the p distinct roots of xp-a, so F splits a separable polynomial, and is galois. The size of the galois group G is the dimension of F/K, which is the degree of our irreducible polynomial, which is p. The only group of order p is cyclic, hence F/K is a cyclic extension of order p.
Conversely, let F/K be a cyclic extension of order p. Let r be a primitive pth root of 1 as described above.
If a field, any field, contains rj for some nonzero j, raise this to the m, where m is the inverse of j mod p. The result is r. Thus our (generic) field contains all the powers of r. One conjugate brings in the rest.
If F contains r, and r is not in K, then K(r) is an intermediate extension of dimension p-1 inside F. Since F/K has dimension p, this is impossible. So if r is in F, it lies in K.
If r is not in F then adjoin r to create a larger field E. All the conjugates of r are present, E is a splitting field, and E is normal over F.
Think of E as K adjoin (F and the powers of r). Since F is a splitting field, and the powers of r form another splitting field, E is a splitting field over K. The associated polynomials are separable, hence E/K is galois.
The galois group of E/F is cyclic, generated by the automorphism r → rj, where j is primitive mod p.
Let's try to get a handle on the galois group of E/K. Let c generate the cyclic galois group of F/K. This is an automorphism on F, and it can be extended to a field homomorphism from F(r) onto F(rj), since r and rj are conjugates. (Remember, j is a primitive root mod p.) Now this homomorphism is really an automorphism on E/K. Call this automorphism d.
What is the order of d inside the galois group? We must invoke d p times before F gets back to where it started. Thus the order of d is divisible by p. At the same time, the root r must return to its original position. Thus the order of d is divisible by p-1. Since p and p-1 are coprime, |d| = p×(p-1), which is the order of the group. The galois group is cyclic.
What is the norm of r in E/K? Let L = K(r), and take it step by step. The norm of r in L/K is the product of the conjugates, which is 1. Raise this to the p power to get |r| in E/K.
Since |r| = 1, apply the trace/norm theorem, and there is a nonzero w in E with w/d(w) = r. Since d(w) is not equal to w, w does not lie in K. Let u = 1/w, so that d(u) = ur.
Notice that d fixes up, hence up lies in K. In other words, u is the pth root of a, for some a in K.
The subgroup generated by the automorphism dp has order p-1 in the galois group G, and fixes an intermediate extension of dimension p. Yet d is based on c, and cp fixes F. So dp fixes a field of dimension p, and it fixes F, hence it fixes precisely F, and nothing more. Since dp fixes u, u is in F.
The field K(u) cannot be intermediate between K and F, since the dimension of F/K is prime; therefore K(u) = F. The polynomial xp-a is irreducible over K.
Since F is galois, it splits the separable polynomial xp-a. This means F contains all the conjugates of u, which are all distinct. Take the quotient of any two conjugates and raise this to the p power. The result is 1. Thus F contains a pth root of 1. If F contains r then K contains r (we showed this earlier). So K contains r after all. We didn't need to "adjoin" r; it was there all the time.
In summary, let K be a field whose characteristic is not p. The extension F/K is cyclic of order p, iff K contains a primitive pth root of 1, and F = K(u), where u is a root of the irreducible polynomial xp-a, for some a in K.