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

By Stein W.A.

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If no qi is contained in p1 , then for each i there is an ai ∈ qi such that ai ∈ p1 . But the product of the ai is in p1 · · · pn , which is a subset of p1 , which contradicts that p1 is a prime ideal. Thus qi = p1 for some i. We can thus cancel qi and p1 from both sides of the equation by multiplying both sides by the inverse. Repeating this argument finishes the proof of uniqueness. 12. If I is a fractional ideal of OK then there exists prime ideals p1 , . . , pn and q1 , . . , qm , unique up to order, such that I = (p1 · · · pn )(q1 · · · qm )−1 .

For example, the field √ of fractions of √ Z is (canonically isomorphic to) Q and the field of fractions of Z[(1 + 5)/2] is Q( 5). 19). 1. We compute the fraction fields mentioned above. sage: Frac(ZZ) Rational Field In Sage the Frac command usually returns a field canonically isomorphic to the fraction field (not a formal construction). 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.

Given a prime p that we wish to factor in OK , we first find a p-maximal order O. 3 to find the prime ideals p of O that contain p. Second, we find the exponents e such that pe exactly divides pO. The resulting factorization in O completely determines the factorization of pOK . A p-maximal order can be found reasonably quickly in practice using algorithms called “round 2” and “round 4”. To compute OK , given an order Z[α] ⊂ OK , one takes a sum of p-maximal orders, one for every p such that p2 divides Disc(Z[α]).

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


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