(⇒) (1) All that needs to be done is to prove closure under subtraction and multiplication. We have (a+bfω)−(c+dfω)=(a−c)+(b−d)fω. As well, ω2=D or D−14+1+√D2 depending on D modulo 4, so(a+bfω)(c+dfω)=(ac)+(bc+ad)fω+bdf2ω2is evidently within the ring as well. (2) Clearly 1∈Of; set a=1 and b=0. (3) Construct a mapping φ:[O]→Z×Z by a+bω↦(a,b). This is well defined by unique a,b form of the argument, clearly surjective and injective, andφ((a+bω)+(c+dω))=φ((a+c)+(b+d)ω)=(a+c,b+d)=(a,b)+(c,d)=φ(a+bω)+φ(c+dω)so that φ is homomorphic and now an isomorphism. So [O]≅Z×Z and by definition we can tell [Of]≅φ([Of])=Z×fZ. By 5.1.14, we have (Z×Z)/(Z×fZ)≅(Z/Z)×(Z/fZ)≅Z/fZ is a group of order f, so that f=|Z×Z:Z×fZ|=|φ([O]):φ([Of])|=|[O]:[Of]|.
Lemma 2: Any nontrivial subgroup of Z is of the form nZ for some n∈Z+. Proof: Let distinct a,b be chosen with n=|a−b| minimal; we thus have nZ is contained within the subgroup. Assume there is an element x not within nZ: We have x=qn+r with 0<r<n, and since qn∈nZ, we have x−qn<n=|a−b|, a contradiction. So nZ is precisely the subgroup in question. ◻
(⇐) R's identity must be 1, as n(a+bfω)≠a+bfω for any n≠1 and a≠0≠b. Therefore, we have 1Z×0=Z×0≤φ([R]). For any (a,b)∈φ([R]), we also have (a,0)∈φ([R]), so that (0,b)∈φ([R]), implying the second factor of φ([R]) is a subgroup of Z and therefore of the form nZ. Since |Z×Z:Z×nZ|=n, we have n=f and now φ([R])=Z×fZ=φ([Of]) and now [R]=[Of]. Since the specifications of multiplication are predefined in C, we have R=Of. ◻
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