Algebraic number theory (Math 784) by Filaseta M.

By Filaseta M.

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Then β|k1 k2 implies that β|k1 and β|k2 , contradicting the minimality of k. Hence, k is prime. Suppose now that β|p and β|q in R where p and q are distinct rational primes. Then using that there exist rational integers x and y such that px + qy = 1, we deduce that β|1, a contradiction. Hence, there is a unique rational prime p such that β|p in R. 43 Homework: For the following problems, take R to be the ring of algebraic integers in an algebraic number field Q(α). (1) Prove that if u is a unit in R and β is an irreducible element of R, then uβ is irreducible.

Comment: The polynomial F (x) in Theorem 37 is called the field polynomial for β. A lemma about conjugates that we will use momentarily is: 33 Lemma. Given the notation above, let w(x) ∈ Q[x] with β = w(α) (we do not require deg w ≤ n − 1). Then, for each j ∈ {1, 2, . . , n}, the field conjugate βj = h(αj ) satisfies βj = w(αj ). Proof. Divide w(x) by f (x) (the minimal polynomial for α) to get w(x) = f (x)q(x) + r(x) where q(x) and r(x) are in Q[x] with r(x) ≡ 0 or deg r ≤ n − 1. Then β = w(α) = r(α) so that, in fact, r(x) = h(x).

Hint: Consider two cases as in the proof above. Use arithmetic in Z[ −2]. ) (2) Consider all the pairs of integers satisfying y 2 + 11 = x3 . (a) Prove that there are no solutions with y odd. (b) Prove that there are ≤ 100 pairs of integers (x, y) with y 2 + 11 = x3 . • A conjecture of Ramanujan. The next result was conjectured by Ramanujan and first verified by Nagell. Theorem 64. The only solutions of the equation x2 + 7 = 2n where x and n are in Z are given by x = ±1, ±3, ±5, ±11, and ±181 and n = 3, 4, 5, 7, and 15, respectively.

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