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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