Wednesday, June 5, 2013
Primary Decomposition Theorem (10.3.18)
Dummit and Foote Abstract Algebra, section 10.3, exercise 18:
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Let M be an R-module for R a PID, and let Ann(M)=(a)≠0. Let a=pα11...pαkk be its decomposition into prime factors. Let Mi=Ann(pαii), i.e. the pi-primary component of M. ProveM=M1⊕ ... ⊕Mk
Proof: Let ni=a/pαii. We can see that no prime factor divides all of the ni, so that their greatest common divisor is 1 and there exist ri such that 1=r1n1+...+rknk. For any m∈M we have m=r1n1m+...+rknkm, and since pαii(rinim)=ri(am)=0, we have rinim∈Mi and now M=M1+...+Mk. To observe uniqueness of sums, letm∈Mj∩(M1+...+Mj−1+Mj+1+...+Mk)with Ann(m)=(d). We have d∣pαjj from the first half as well as d∣nj from the second so that d=1 and now 1m=m=0. ◻
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