I have a question on the following proof, which takes place over an R-module M where R is a PID. Here v∈M, 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)=α1⋯αn, where the αi's are pairwise coprime, then v has form v=u1+⋯+un where o(ui)=αi.
Proof: Let μ=α1⋯αn. The scalars βk=μ/αk are coprime, so there exist ai∈R such that
a1β1+⋯+anβn=1
Then
v=(a1β1+⋯+anβn)v=a1β1v+⋯+anβnv.
Since o(βkv)=μ/gcd and since a_k and \alpha_k are relatively prime, we have o(\alpha_k\beta_kv)=\alpha_k.
My question: How do we know a_k and \alpha_k are relatively prime?
Answer
Because \alpha_k divides \beta_i for all i \neq k. If some factor of \alpha_k divided a_k (hence a_k\beta_k) it would divide all the a_i\beta_i; hence the expression
a_1\beta_1+\cdots+a_n\beta_n=1
would not be possible.
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