By Stein W.A.

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**Extra info for Algebraic number theory, a computational approach**

**Example text**

Sage: I*J Fractional ideal (-1/2*a - 3/2) of Number Field ... Since fractional ideals I are finitely generated, we can clear denominators of a generating set to see that there exists some nonzero α ∈ K such that αI = J ⊂ OK , with J an integral ideal. Thus dividing by α, we see that every fractional ideal is of the form aJ = {ab : b ∈ J} for some a ∈ K and integral ideal J ⊂ OK . For example, the set 12 Z of rational numbers with denominator 1 or 2 is a fractional ideal of Z. 8. The set of fractional ideals of a Dedekind domain R is an abelian group under ideal multiplication with identity element R.

Let b1 = (a0 , a1 , . . , ad+1 ) be the first row of B and notice that B is obtained from A by left multiplication by an invertible integer matrix. Thus a0 , . . 1) that equals ad+1 . Moreover, since B is LLL reduced we expect that ad+1 is relatively small. 3. Output f (x) = a0 + a1 x + · · · ad xd . We have that f (α) ∼ ad+1 /K, which is small. Thus f (x) may be a very good candidate for the minimal polynomial of β (the algebraic number we are approximating), assuming d was chosen minimally and α was computed to sufficient precision.

If a = c + d −6, then Norm(a) = c2 + 6d2 ; since the equation c2 + 6d2√= 2 has no solution with√c, d ∈ Z, there is no element in OK with norm 2, so −6 is irreducible. Also, −6 is not a unit times 2 or times 3, since again the norms would not match up. Thus 6 can not be written uniquely as a product of irreducibles in OK . 12, however, implies that the principal ideal (6) can, however, be written uniquely as a product of prime ideals. 1) √ √ where each of the ideals (2, 2 + −6) and (3, 3 + −6) is prime.

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

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Categories: Algebraic Geometry