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?
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.
No comments:
Post a Comment