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Thursday, April 25, 2013

Order of Conductor f (7.1.23)

Dummit and Foote Abstract Algebra, section 7.1, exercise 23:

MathJax TeX Test Page Define Of=Z[fω]={a+bω | a,bZ}. Letting [R] denote the abelian group in the ring R, prove:R=OfR subring of O1R|[O]:[R]|=f Proof: Lemma 1: For HG groups and φ an isomorphism, we have |G:H|=|φ(G):φ(H)|. Proof: Let n1,... be a complete set of coset-distinct representatives for H in G. We claim that their images are a complete set of coset-distinct representatives of φ(H) in φ(G). For any gG, we have g=nih for some index i and hH. Therefore, we have φ(g)=φ(nih)=φ(ni)φ(h). Assuming φ(ni) and φ(nj) are in the same coset of φ(H), we have φ(ni)=φ(nj)φ(h)=φ(njh), so that ni=njh and ni and nj are in the same coset of H, a contradiction. Thus, there are just as many coset representatives of φ(H) in φ(G) as H in G. 

() (1) All that needs to be done is to prove closure under subtraction and multiplication. We have (a+bfω)(c+dfω)=(ac)+(bd)fω. As well, ω2=D or D14+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 1Of; 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 nZ+. Proof: Let distinct a,b be chosen with n=|ab| 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 qnnZ, we have xqn<n=|ab|, a contradiction. So nZ is precisely the subgroup in question. 

() R's identity must be 1, as n(a+bfω)a+bfω for any n1 and a0b. Therefore, we have 1Z×0=Z×0φ([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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