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Sqrarefree Integers, Fields, Conductors and Maximal Ideals

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Let D be a squarefree integer, and let 0 be the ring of integers in the quadratic field Q(sqrtD). For any positive integer f prove that the set Of = Z[fw] = {a + bfw | a, b E Z} is a subring of 0 containing the identity. Prove that [O:Of]= f (index as additive abelian groups). Prove conversely that a subring of 0 containing the identity and having finite index f in 0 (as additive abelian group) is equal to Of. (The ring Of is called the order of conductor f in the field Q(sqrtD). The ring of integers 0 is called the maximal order in Q(sqrtD).

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Part A:

Taking two arbitrary elements of O_f,

x = a + bfω and y = c + dfω

we find that both

x + y = (a+c) + (b+d)fω

is in O_f , and

x*y = (ac+Dbd) + (ad+bc)fω

is in O_f.

Therefore O_f is a sub-ring.
Taking a = 1 and b = 0, we see that O_f contains the ...

Solution Summary

Sqrarefree Integers, Fields, Conductors and Maximal Ideals are investigated. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.