By David Hilbert
This ebook is a translation into English of Hilbert's "Theorie der algebraischen Zahlkrper" most sensible referred to as the "Zahlbericht", first released in 1897, during which he supplied an elegantly built-in assessment of the advance of algebraic quantity idea as much as the tip of the 19th century. The Zahlbericht supplied additionally a company beginning for extra study within the topic. it truly is in keeping with the paintings of the nice quantity theorists of the 19th century. The Zahlbericht should be noticeable because the place to begin of all 20th century investigations in algebraic quantity idea, reciprocity legislation and sophistication box conception. For this English version an creation has been further via F. Lemmermeyer and N. Schappacher.
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Additional info for The Theory of Algebraic Number Fields
12), and the proof of the proposition is complete. 12). 4 of the previous chapter determines the image of δk for k = 1, · · · , p − 1. 1. Let k ≥ 1 be any integer. If k = 1 or k≡1 mod (p−1), then δk (U∞ ) = Zp . If k is greater than 1 and k ≡ 1 mod (p − 1), then δk (U∞ ) = pm Zp , where m = 1 + ordp (k − 1). To prove this proposition, we need an algebraic description of Λ(G) as p − 1 copies of the ring R. 5, which is based on Mahler’s theorem. 13) where is the cyclic group of order p − 1, and Γ is isomorphic to Zp .
1. The restriction to Λ(G)+ of the natural surjection from Λ(G) onto Λ(G) induces an isomorphism Λ(G)+ Λ(G). 2) Proof. 3) and write Gn = Gal(Fn /Q), Gn = Gal(Fn /Q). Let πn : Zp [Gn ] −→ Zp [Gn ] denote the natural surjection. We claim that πn induces an isomorphism from Zp [Gn ]+ onto Zp [Gn ]. Indeed, it is clear that πn is surjective, and that it maps Zp [Gn ]− to zero. To complete the proof, we note that the Zp -rank of Zp [Gn ]+ is equal to ((p − 1)/2)pn , because Zp [Gn ](i) Zp [Gn ]+ = i even i mod p−1 where the sum on the right is taken over the eigenspaces for the even powers modulo (p − 1) of the character giving the action of G0 on µp .
Suppose now that ξ is a pseudo-measure on G with χk (ξ) = 0 for all k > 0. For each u ∈ Z× p , let σu denote the unique element of G with χ(σu ) = u. Now choose u = 1 + p. Then χk (σu − 1) = (1 + p)k − 1 = 0 for all k > 0, so that σu − 1 is not a zero divisor in Λ(G) by the remark above. But the hypothesis χk (ξ) = 0 implies that χk ((σu − 1)ξ) = 0 for all k > 0. As σu − 1 is not a zero divisor, the remark at the end of the previous paragraph shows that ξ = 0. 50 4 Cyclotomic Units and Iwasawa’s Theorem Let Q(G) be the total ring of quotients of Λ(G).
The Theory of Algebraic Number Fields by David Hilbert