I am working towards understanding the statement and proof of the following theorem:
Let $R$ be a PID and $a_1,...,a_r \in R \setminus \{0_R\}$. Then, there are $q_1,...,q_s \in R\setminus \{0_R\}$, such that $q_i$ divides $q_{i+1}$ for all $i$, and $\prod_{i=1}^r R/Ra_i \cong \prod_{i=1}^s R/Rq_i$. Also, $q_s$ is an LCM of $\{a_1,...,a_r\}$
I got stuck on a preceding proposition:
Let $R$ be a PID and $A= \prod_{i=1}^m R/Ra_i$ and $B = \prod_{i=1}^n R/Rb_i$ where $a_1,...,a_m,b_1,...,b_n \in R\setminus \{0_R\}$, and $b_i$ divides $b_{i+1}$ for all $i$. If $A \cong B$, then $b_n$ is an LCM of $\{a_1,...,a_m\}$.
Here is the provided proof:
Let $a$ be an LCM of $\{a_1,...,a_m\}$. (( $aA = 0 \implies aB = 0$, hence $b_i$ divides $a$ for all $i$ )), so $b_n$ divides $a$. Also $b_i$ divides $b_n$ for all $i$, so $b_n B = 0$, so $b_n A = 0$, so $a_i$ divides $b_n$ for all $i$, hence $a$ divides $b_n$. Thus, $a$ and $b_n$ are associated. $\square$
My questions are:
What does $aA$ mean? What is its definition and what structure does it have? Can you elaborate on the highlighted part of the proof?
The notation is a little terse, but it seems $aA$ is the set $\{\bar{a}x\mid x \in A\}$ where, if you want to be specific, $\bar{a}$ is the element $(a+(Ra_1),a+(Ra_2),...,a+(Ra_r)) \in A$.
To see that $aA=0$, look more closely at what coset $a+(Ra_1)$ refers to: it is the image of $a$ under the quotient map $R\to R/a_1R$. From the fact that $a$ is a multiple of $a_1$, this is the same as the image of $a_1$, which is $0$. Because $a$ is a lcm of the $a_i$, $$\bar{a} = (a+(Ra_1),a+(Ra_2),...,a+(Ra_r)) = (0+(Ra_1),0+(Ra_2),...,0+(Ra_r)) = 0 \in A$$and clearly $\{\bar{a}x\mid x \in A\} = \{0x\mid x \in A\} = \{0\}$.