By Robert B. Ash
The writer offers enough information for college students to navigate the problematic proofs of the Dirichlet unit theorem and the Minkowski bounds on point and perfect norms. extra themes comprise the factorization of top beliefs in Galois extensions and native in addition to international fields, together with the Artin-Whaples approximation theorem and Hensel's lemma. The textual content concludes with 3 worthwhile appendixes. aimed toward arithmetic majors, this path calls for a historical past in graduate-level algebra and a familiarity with indispensable extensions and localization.
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Additional info for A Course in Algebraic Number Theory
N−1 of L = Q(α), where the minimal polynomial f of α has degree n. 1. Let c1 , . . , cr1 be the real conjugates of α, that is, the real roots of f , and let cr1 +1 , cr1 +1 , . . , cr1 +r2 , cr1 +r2 be the complex (=non-real) conjugates. Show that the sign of the discriminant is the sign of r2 (cr1 +i − cr1 +i )2 . i=1 2. Show that the sign of the discriminant is (−1)r2 , where 2r2 is the number of complex embeddings. 3. Apply the results to α = ζ, where ζ is a primitive (pr )th root of unity.
CYCLOTOMIC EXTENSIONS Lemma For every positive integer m, we have Z[ζ] + pm B = B. Proof. We ﬁrst prove the identity with p replaced by π. If b ∈ B, then b + (π) = t + (π) for some integer t, hence b−t ∈ (π). Thus Z[ζ]+πB = B, and consequently πZ[ζ]+π 2 B = πB. Now iterate: If b ∈ B, then b = b1 + b2 , b1 ∈ Z[ζ], b2 ∈ πB. Then b2 = b3 + b4 , b3 ∈ πZ[ζ] ⊆ Z[ζ], b4 ∈ π 2 B. Observe that b = (b1 + b3 ) + b4 , so Z[ζ] + π 2 B = B. Continue r in this fashion to obtain the desired result. 3), π ϕ(p ) is p times a unit, so if r m m = ϕ(p ), we can replace π B by pB, so that Z[ζ] + pB = B.
If ζ were a root of an irreducible factor of Φpr , then the degree of the cyclotomic extension would be less than ϕ(pr ), contradicting what we have just proved. 5 Lemma Let B be the ring of algebraic integers of Q(ζ). Then (π) is a prime ideal (equivalently, π is a prime element) of B. The relative degree f of (π) over (p) is 1, hence the injection Z/(p) → B/(π) is an isomorphism. Proof. If (π) were not prime, (p) would have more than ϕ(pr ) prime ideal factors, which is impossible, in view of the ram-rel identity.
A Course in Algebraic Number Theory by Robert B. Ash