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Discriminant(); D 2945785 sage: factor(D) 5 * 353 * 1669 The order Z[a] has the same discriminant as f (x), which is the same as the discriminant of OK , so Z[a] = OK and we can apply the above theorem. = QQ[] sage: discriminant(x^5 + 7*x^4 + 3*x^2 - x + 1) 2945785 We have x5 + 7x4 + 3x2 − x + 1 ≡ (x + 2) · (x + 3)2 · (x2 + 4x + 2) (mod 5), which yields the factorization of 5OK given before the theorem. If we replace a by b = 7a, then the index of Z[b] in OK will be a power of 7, which is coprime to 5, so the above method will still work.

41 42CHAPTER 3. , a subring of OK of finite index – is also the number field again. 2 (Integrally Closed). An integral domain R is integrally closed in its field of fractions if whenever α is in the field of fractions of R and α satisfies a monic polynomial f ∈ R[x], then α ∈ R. For example, every field√is integrally closed in its field of fractions, as is the ring Z of √ integers. 3. If K is any number field, then OK is integrally closed. Also, the ring Z of all algebraic integers (in a fixed choice of Q) is integrally closed.

Proof. Suppose α ∈ OK ∩ Q with α = a/b ∈ Q in lowest terms and b > 0. Since α is integral, Z[a/b] is finitely generated as a module, so b = 1. To prove that QOK = K, suppose α ∈ K, and let f (x) ∈ Q[x] be the minimal monic polynomial of α. , the polynomial obtained from f (x) by multiplying the coefficient of xdeg(f ) by 1, multiplying the coefficient of xdeg(f )−1 by d, multiplying the coefficient of xdeg(f )−2 by d2 , etc. If d is the least common multiple of the denominators of the coefficients of f , then the minimal monic polynomial of dα has integer coefficients, so dα is integral and dα ∈ OK .

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Algebraic number theory, a computational approach by Stein W.A.


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