By Takashi Ono (auth.)
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Since the relation k(a, f3):::) key) is obvious, we have only to prove k(a, f3) c key). To do this, put g(X) = fa(Y - eX); note that g and fp are polynomials in k(y)[X]. Now, we have g(fJ) = fa(Y - ef3) = fa(a) = 0 and ffJ(f3) = 0; this means that X - fJ is a common divisor of g, fp in Q[X]. , e(f3 - fJj) = ai = a, contradicting the choice of e. Therefore we proved that (g, fp) = X - f3. , f3 E key). Finally, we have a = y - ef3 E key). Hence k(a, f3) c key). D. 6. Let K / k be a finite extension of algebraic number fields.
Let e = (n, p -1), n EN, and write p -1 = ef. Then, we have xn == a (p) has a solution ¢:> af == 1 (p ). When the equation has a solution, the number of solutions (modp) is e. PROOF. 20, to solve xn == a (p) is equivalent to solving ny == b (p - 1), a linear equation, where y = indr x, b = indr a. 8). If that is so, dividing both sides of ny == b (p -1) by (e), the equation n'y == b' (f), n = en', b = eb', has a unique solution Yo modulo f. Then e numbers Yo, Yo + f, ... ,Yo + (e -l)f make up solutions of ny == b (p -1).
22. An integer a, p fa, is called an nth power residue or an nth power nonresidue according as the equation xn == a (p) has a solution or not. if p is an odd prime, an integer a, p fa, is called a quadratic residue or a quadratic nonresidue according as the equation x 2 == a (p) has a solution or not. 21 with n = 2, p =fo 2, we see that, for a E Z, p f a, x 2 == a (p) has a solution ¢:> at == 1 (p), 2f=p -1. In other words, we have the following proposition. 17. Let a be an integer and p be an odd prime such that p f a.