By Filaseta M.

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**Additional info for Algebraic number theory (Math 784)**

**Sample text**

We deduce α = G(β)/F (β) ∈ Q(β), and the theorem follows. 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.

Since q ≡ ±2 (mod 5), we obtain a contradiction as ±3 are not squares modulo 5. √ 2 ≡ 5 (mod q) √ for some integer a. Then q prime in R and q|(a + √ 5)(a − √ Now, assume a √ 5) implies q|(a + 5) or q|(a − 5). This gives a contradiction as neither (a + 5)/q nor √ (a − 5)/q are in R (as q > 2). Thus, there are no solutions to the congruence x2 ≡ 5 (mod q). Lemma 2. Let q be a rational prime with q = 2 and q ≡ ±2 (mod 5). Then 5(q−1)/2 ≡ −1 (mod q). Proof. From Lemma 1, there are no solutions to the congruence x2 ≡ 5 (mod q).

M be the distinct roots of the minimal polynomial g(x) for β. Note that for i ∈ {1, 2, . . , n} and j ∈ {2, 3, . . , m}, there exists a unique x = x(i, j) such that αi + xβj = α + xβ. It follows that there is a rational number c for which the number γ = α + cβ satisfies γ = αi + cβj for all i ∈ {1, 2, . . , n} and j ∈ {2, 3, . . , m}. We prove Q(γ) = Q(α, β). To prove Q(γ) = Q(α, β), we show that γ ∈ Q(α, β) and that each of α and β is in Q(γ). Clearly, γ ∈ Q(α, β). Let h(x) = f (γ − cx) ∈ Q(γ)[x].