The following is the part of theorem 6.4 of the book Advanced linear algebra of Steven Roman related to my question:
Context: here $R$ is a principal ideal domain and $M$ an $R$-module. Also, for $v\in M$ we set $o(v)$ as the order of the submodule generated by $v$. The order is any element of the annihilator of $v$, that is, any generator of the annihilator of $v$.
Theorem: for some $v\in M$ suppose that $o(v)=\alpha_1\cdots\alpha_n$ where the $\alpha_i's$ are pairwise relatively prime, then $v$ has the form $$v=u_1+\ldots+u_n$$ where $o(u_i)=\alpha_i$.
In some previous place Steven had set $\mu:=\alpha_1\cdots\alpha_n$ and the proof of the above theorem start saying
The scalars $\beta_k:=\mu/\alpha_k$ are relatively prime and so there exists $a_k\in R$ such that $$a_1\beta_1+\ldots+a_n\beta_n=1$$
Since $o(\beta_k v)=\mu/\gcd(\mu,\beta_k)=\alpha_k$ and since $a_k$ and $\alpha_k$ are relatively prime then $o(a_k\beta_k v)=\alpha_k$.
I follow everything except the statement that $a_k$ and $\alpha_k$ are relatively prime, why? I can't see it. Can someone enlighten me here please?
Note that $\alpha_k\mid\beta_j$ for $k\neq j$. So any common divisor of $a_k$ and $\alpha_k$ divides $$ a_1\beta_1+\cdots+a_n\beta_n=1, $$ since it divides $a_k$ and hence $a_k\beta_k$, and divides $a_j\beta_j$ for $j\neq k$ since it divides $\beta_j$. Hence it is a divisor of $1$ so $\gcd(a_k,\alpha_k)=1$.