Saturday, 12 July 2014

abstract algebra - Question on decomposition of cyclic modules.



I have a question on the following proof, which takes place over an R-module M where R is a PID. Here vM, and o(v) is the order of v, defined to be a generator (or any of its associates) of the annihilator of v in R.




If o(v)=α1αn, where the αi's are pairwise coprime, then v has form v=u1++un where o(ui)=αi.



Proof: Let μ=α1αn. The scalars βk=μ/αk are coprime, so there exist aiR such that
a1β1++anβn=1


Then
v=(a1β1++anβn)v=a1β1v++anβnv.


Since o(βkv)=μ/gcd(μ,βk)=αk and since ak and αk are relatively prime, we have o(αkβkv)=αk.



My question: How do we know ak and αk are relatively prime?


Answer



Because αk divides βi for all ik. If some factor of αk divided ak (hence akβk) it would divide all the aiβi; hence the expression
a1β1++anβn=1


would not be possible.


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