# Download PDF by Paula Ribenboim: Algebraic Numbers (Pure & Applied Mathematics Monograph) By Paula Ribenboim

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1) Proposition. The mappings are mutually inverse 1-1 -correspondences between the prime ideals q 5 A \ S of A and the prime ideals 8 of AS-'. Chapter I. Algebraic Integers 66 Proof: If q 5 A I \ 1%1 q E q, s E S) 0 0' is a prime ideal of A S - ' . d = implies that s"aal = qss' E q . Therefore aa' E q , because ss' S" ' s" \$! Furthermore one has q=OnA, since = a E 12 f l A implies q = a s E q , whence a E q because s \$! q. Conversely, let 12 be an arbitrary prime ideal of AS-'. Then q = 12 fl A is obviously a prime ideal of A , and one has q g A \ S.

As the characteristic p of o / p does not divide n, the polynomials Xn - 1 and nxn-' have no common root in alp. So Xn - 1 mod p has no multiple roots. We therefore see that passing to the quotient o -+ o / p maps the group pn of n-th roots of unity bijectively onto the group of n-th roots of unity of alp. In particular, the primitive n-th root of unity ( modulo p remains a primitive n-th root of unity. The smallest 'extension field of IFp = Z / p Z containing it is the field F p f p ,because its multiplicative group P* is cyclic of order p f p - l .

It is of the most beautiful simplicity. 3) Proposition. Let n = pVp be the prime factorization of n and, for every prime number p, Jet fp be the smallest positive integer such that pfp E 1 mod n/pVp Then one has in Q (0the factorization Multiplying this by h and substituting the result ho = h20 obtain x 2 0 Z[(] = 0. + + h Z [ < ] ,we p = ( p , . . pr)"(pYP), where p , , . . , pr are distinct prime ideals, all of degree ,f,,. Iterating this procedure, we find h t o + Z [ ( ] = o for all t 2 1 .