By Takashi Ono (auth.)

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**Example text**

27 is certainly useful to determine whether the equation x 2 == a (p) has a solution or not. , the algebraic number theory. 20 already hint at this. 45. Let q be a prime such that q == 1 (8) and a be an integer such that p2 l' a for any prime p and that X4 == a (q) has no solution in 7L. Prove that the equation X4 - qy4 = az 2 has no integral solution other than x = y = z = o. Also find a pair {q, a} which satisfies the above condition. 2 Basic Concepts of Algebraic Number Fields In Chapter 2, we shall extend the results of Chapter 1 (number theory in 10) to the case of an algebraic number field K.

A/ are relatively prime ¢::>the equation AIxI + ... + A/x/ = 1 has a solution Xi E K[X], 1:5 i :51. 4*. Let A, B, C be polynomials in K[ X] such that (A, B) = 1. Then A I BC=>A I c. 5*. Let P be irreducible. P lAB P IA or P I B. 6*. Every polynomial in K[X] can be written as a product of irreducible polynomials; it can be so written in one and only one way except for the order of factors and for multiplication by elements of KX. 20. 2* show that the ring R = Z or K[X] is "euclidean". , K = IFp = Z/pZ.

D. duces the group isomorphism R X =Rf x· .. x RT'. 36. 15 is surjective. 10. If(n, m) f in the proof of = 1 then cp(nm) = cp(n)cp(m). PROOF. 15 with 1=2. D. 16. When m = P'1! ) ... 15. Therefore the function cp is determined if we find cp(pe), e ~ 1. When e = 1, we find cp(p) = p - 1 because 7L/p7L is a field. 11. If P is a prime, we have q;(pe) = pe-l(p 1), e;::: 1. PROOF. Put G = (7L/pe+17L), G' = (7L/p e7L), and consider a homomorphism f: G~ G' defined by (f[X]p<+I) = [x]p" f is surjective, because if [X]pe E G' (hence p f x) we have f([X]pe+l) = [X]pe.