By Stein W.A.
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Extra info for Algebraic number theory, a computational approach
4. 3 implies that the norm and trace down to L of α is an element of OL , because the sum and product of algebraic integers is an algebraic integer. 5. 5. Let K be a number field. , QOK = K and OK is an abelian group of rank [K : Q]. Proof. 19 that QOK = K. Thus there exists a basis a1 , . . , an for K, where each ai is in OK . Suppose that as x = ni=1 ci ai ∈ OK varies over all elements of OK the denominators of the coefficients ci are not all uniformly bounded. Then subtracting off integer multiples of the ai , we see that as x = ni=1 ci ai ∈ OK varies over elements of OK with ci between 0 and 1, the denominators of the ci are also arbitrarily large.
DEDEKIND DOMAINS AND UNIQUE FACTORIZATION OF IDEALS To finish the proof that p has an inverse, we observe that d preserves the finitely generated OK -module p, and is hence in OK , a contradiction. More precisely, if b1 , . . , bn is a basis for p as a Z-module, then the action of d on p is given by a matrix with entries in Z, so the minimal polynomial of d has coefficients in Z (because d satisfies the minimal polynomial of d , by the Cayley-Hamilton theorem – here we also use that Q ⊗ p = K, since OK /p is a finite set).
As we will see, in general the problem of computing OK given K may be very hard, since it requires factoring a certain potentially large integer. 3. 17 (Order). An order in OK is any subring R of OK such that the quotient OK /R of abelian groups is finite. ) As noted above, Z[i] is the ring of integers of Q(i). For every nonzero integer n, the subring Z+niZ of Z[i] is an order. The subring Z of Z[i] is not an order, because Z does not have finite index in Z[i]. Also the subgroup 2Z + iZ of Z[i] is not an order because it is not a ring.
Algebraic number theory, a computational approach by Stein W.A.