Number Theory

Problems in Algebraic Number Theory by M. Ram Murty, Jody (Indigo) Esmonde

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By M. Ram Murty, Jody (Indigo) Esmonde

This can be a very precious e-book for somebody learning quantity idea. it is specially necessary for amatuer mathematicians studying on their lonesome. This one is equal to the older variation with extra tricks and extra distinctive clarification. yet would it BE nice to go away a bit room for the readers to imagine on their own?! you'll gain the convenience from considering challenging in addition to operating demanding!

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For N ≥ n, suppose this holds ∀αj , j < N . αN = αN −n αn = αN −n [−(a0 + a1 α + · · · + an−1 αn−1 )] = (−αN −n a0 )1 + (−αN −n a1 )α + · · · + (−αN −n an−1 )αn−1 . By our inductive hypothesis, −αN −n ai ∈ Z[α] ∀i = 0, 1, . . , n − 1. Then Z[α] is a Z-module generated by {1, α, . . , αn−1 }. 3. ALGEBRAIC NUMBER FIELDS 37 (c) ⇒ (d) Let M = Z[α]. Clearly αZ[α] ⊆ Z[α]. (d) ⇒ (a) Let x1 , . . , xr generate M over Z. So M ⊆ Zx1 + · · · + Zxr . By assumption αxi ∈ M ∀i = 1, . . , r. It follows that there exists a set of n cij ∈ Z such that αxi = j=1 cij xj ∀i = 1, .

So gcd(φ(x), g(x)) = c(x − β), c ∈ C∗ . Then c(x − β) ∈ L[x] which implies that c, cβ ∈ L, and so β ∈ L. Now, θ = α + λβ ∈ L which means that α ∈ L. So Q(α, β) ⊆ L = Q(θ). Thus, we have the desired equality: Q(α, β) = Q(θ). ✷ This theorem can be generalized quite easily using induction: for a set α1 , . . , αn of algebraic numbers, there exists an algebraic number θ such that Q(α1 , . . , αn ) = Q(θ). Therefore, any algebraic number field K is Q(θ) for some algebraic number θ. 3 Let α be an algebraic number and let p(x) be its minimal polynomial.

Thus, m is maximal, ⇔ R/m has no nontrivial ideals, ⇔ R/m is a field. 53 54 CHAPTER 5. DEDEKIND DOMAINS (b) ℘ is a prime ideal ⇔ ab ∈ ℘ ⇒ a ∈ ℘ or b ∈ ℘, ⇔ ab + ℘ = 0 + ℘ in R/℘ ⇒ a + ℘ = 0 + ℘ or b + ℘ = 0 + ℘ in R/℘, ⇔ ⇔ R/℘ has no zero-divisors, R/℘ is an integral domain. (c) Suppose that ℘ ⊇ ab, ℘ ⊇ a. Let a ∈ a, a ∈ / ℘. We know that ab ∈ ℘ for all b ∈ b since ab ⊆ ℘. But, a ∈ / ℘. Thus, b ∈ ℘ for all b ∈ b, since ℘ is prime. Thus, b ⊆ ℘. (d) (By induction on r). The base case r = 1 is trivial.

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