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I have a question on the following proof, which takes place over an $R$-module $M$ where $R$ is a PID. Here $v\in M$, 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 $o(v)=\alpha_1\cdots\alpha_n$, where the $\alpha_i$'s are pairwise coprime, then $v$ has form $v=u_1+\cdots+u_n$ where $o(u_i)=\alpha_i$.

Proof: Let $\mu=\alpha_1\cdots\alpha_n$. The scalars $\beta_k=\mu/\alpha_k$ are coprime, so there exist $a_i\in R$ such that $$ a_1\beta_1+\cdots+a_n\beta_n=1 $$ Then $$ v=(a_1\beta_1+\cdots+a_n\beta_n)v=a_1\beta_1v+\cdots+a_n\beta_nv. $$ Since $o(\beta_kv)=\mu/\gcd(\mu,\beta_k)=\alpha_k$ 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?

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up vote 1 down vote accepted

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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Thank you Jim. () – Georg Carlin Feb 19 '13 at 5:10

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