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Wednesday, June 5, 2013

Primary Decomposition Theorem (10.3.18)

Dummit and Foote Abstract Algebra, section 10.3, exercise 18:

MathJax TeX Test Page 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 mM we have m=r1n1m+...+rknkm, and since pαii(rinim)=ri(am)=0, we have rinimMi and now M=M1+...+Mk. To observe uniqueness of sums, letmMj(M1+...+Mj1+Mj+1+...+Mk)with Ann(m)=(d). We have dpαjj from the first half as well as dnj from the second so that d=1 and now 1m=m=0. 

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