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By Ash R.B.

It is a textual content for a uncomplicated direction in algebraic quantity conception.

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Extra resources for A Course In Algebraic Number Theory

Example text

If we wish to factor the ideal (2) = 2B of B, the idea is to√factor x2 + 5 mod 2, and the result √ is x2 + 5 ≡ (x + 1)2 mod 2. Identifying x with −5, we form the ideal P2 = (2, 1 + −5), which turns out to be prime. The desired factorization is (2) = P22 . This technique works if B = Z[α], where √ the number field L is Q( α). √ 1. Show that 1 − −5 ∈ P2 , and conclude that 6 ∈ P22 . 2 2. Show that 2 ∈ P22 , hence . √ (2) ⊆ P2 √ 2 3. Expand P2 = (2, 1 + −5)(2, 1 + −5), and conclude that P22 ⊆ (2).

For our group G, even more is true. 6 Proposition The group G consists exactly of all the roots of unity in the field L. Proof. 5), every element of G is a root of unity. Conversely, suppose xm = 1. Then x is an algebraic integer (it satisfies X m − 1 = 0) and for every i, |σi (x)|m = |σi (xm )| = |1| = 1. Thus |σi (x)| = 1 for all i, so log |σi (x)| = 0 and x ∈ G. 7 Proposition B ∗ is a finitely generated abelian group, isomorphic to G × Zs where s ≤ r1 + r2 . Proof. 3), λ(B ∗ ) is a discrete subgroup of Rr1 +r2 .

Now B is the integral closure of A in L, so B is the integral closure of A in S −1 L = L. ] We have now reduced to the PID case already analyzed, and [B /P B : A /P A ] = n. g Now P B = i=1 Piei , and Pi is a nonzero prime ideal of B not meeting S. 2). 6), we have the factorization P B = i=1 (Pi B )ei . By the PID case, g n = [B /P B : A /P A ] = ei [B /Pi B : A /P A ]. i=1 We are finished if we can show that B /Pi B ∼ = B/Pi and A /P A ∼ = A/P . The statement of the appropriate lemma, and the proof in outline form, are given in the exercises.

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