By Serge Lang
This is a moment version of Lang's recognized textbook. It covers all the uncomplicated fabric of classical algebraic quantity conception, giving the coed the historical past useful for the research of additional issues in algebraic quantity thought, resembling cyclotomic fields, or modular forms.
"Lang's books are continuously of significant price for the graduate pupil and the examine mathematician. This up-to-date version of Algebraic quantity concept is not any exception."―-MATHEMATICAL REVIEWS
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Additional info for Algebraic Number Theory
Let a be a fractional ideal of A. Then NTc(aB) = Proof. Immediate. Il[L:Kl. 26 ALGEBRAIC INTEGERS [1, §7] Corollary 2. Assume that L is Galois over K. Then all the e'ß are equal to the same number e (jor 'I3lp), all the h are equal to the same number f (jor 'I3lp), and if then efr = [L:K]. Proof. All the '13 lying above p are conjugate to each other, and hence all the ramification indices and residue class degrees are equal. The last formula is clear. Corollary 3. Assume again that L is Galois over K with group G, and let '13 be a prime of B Zying above p in A.
We denote by Np the number of elements in A/p. We now have two absolute values determined by p. On the one hand the unique absolute value such that und [II, §1] DEFINITIOKS AND COMPLETIOKS 35 and on the other hand the unique absolute value such that 111l"llp = 1 Np . For any a E K, a ,e 0, we have Similarly, if L is a finite extension of K, and Iß lies above p in the ring of algebraic integers B of L, let TI be an element of order 1 at Iß. Then pB = 1ße(~/P) ••• and The fact that ramification indices and residue class degrees are multiplicative in towers insures the consistency of these definitions when we go to finite extensions.
Let e be the ramification index and write where eo is prime to p. Let TI be an element of order 1 at ~, and let ß = TIpT. By the lemma, K(ß) contains a tamely ramified subextension of ramification index eo. The composite of this extension with the maximal unramified subfield of E gives us a tamely ramified extension F of K, and from the definition of ß, it follows that the ramification index of E over F is pT. On the other hand, E is totally ramified over F (because F contains Eu), and hence [E: F] = pT.