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Tuesday, April 30, 2013

Nilpotency and the Augmentation Ideal (7.4.3)

Dummit and Foote Abstract Algebra, section 7.4, exercise 3:

MathJax TeX Test Page (a) Let G be an abelian group of order pn. Prove Aug(FpG)=N(FpG).
(b) Let G={g1,...,gn} be a group and assume R is commutative. Prove rAug(RG)r(g1+...+gn)=0.

Proof: (a) () By the preceding exercise, Aug(FpG) is generated by {g1 | gG}, so for arbitrary xAug(FpG) write x=gGag(g1) for some agFpG for all gG. Prove (g1)pn=0 and is thus nilpotent for any gG, so that by the idealness of N(FpG) we have x is nilpotent. By the binomial theorem for commutative rings, since pn divides (pnk)=pn!k!(nk)! when 0<k<pn, we can observe (g1)pn=gpn1=11=0. () Since FpG/Aug(FpG)Fp is a field, we have Aug(FpG) is a maximal ideal, so that N(FpG){Aug(FpG),FpG}, and is evidently not the latter as 1FpG is not nilpotent. 

(b) Once again write r=r(g1) for some rRG, and recalling that the action of multiplication on a group by one of its elements is a permutation of that group's elements, we observe r(g1+...+gn)=r(g1)(g1+...+gn)=r(g(g1+...+gn)(g1+...+gn))=r((g1+...+gn)(g1+...+gn))=r0=0 

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