Note: This might end up being a question about a simple concept that I forgot about (I am very tired at the time of writing this, after all), so maybe try skipping to the bottom.
I'm learning about modules over a PID in abstract algebra (Nicholson) and there's this theorem (7.2.5, if for whatever reason you want to know):
Let $R$ be a PID, let $p \in R$ be a prime, and let $M$ be a finitely generated, nonzero, $p$-module over $R$. Then there is a decomposition $$ M = Rx_1 \oplus \dots \oplus Rx_t $$ where $o(x_i) = p^{m_i}$, with $m_1 \geq m_2 \geq \cdots \geq m_t \geq 1$. Furthermore, the integers $t, m_1, m_2, \dots, m_t$ are uniquely determined by $M$. More generally, if $K \subseteq M$ is any submodule and $K = Ry_1 \oplus \dots \oplus R y_u$ where $o(y_i) = p^{k_i}$ with $k_1 \geq \cdots \geq k_u \geq 1$, then $u \leq t$ and $k_i \leq m_i$ for each $i=1,\dots,u$.
I'm not confused about the theorem, although it is relevant. It basically says that given such a direct sum, the exponent of the orders of the $x_i$ uniquely determine $M$ given the ring $R$. Hence, we have the following definitions:
- The $t$-tuple $(m_1, \dots, m_t)$ is called the type of the module $M$
- The elements $p^{m_1}, \dots, p^{m_t}$ are called the elementary divisors of $M$.
But then the author says that given a sequence of integers $m_1 \geq m_2 \geq \cdots \geq m_t$ the module $$ R/Rp^{m_1} \oplus \cdots \oplus R/Rp^{m_t} $$ is of type $(m_1, \dots, m_t)$. Why? In the definition, $M$ is the sum of principal ideals, but not here - it's just quotient rings. I feel like I'm missing something very obvious, can someone help me out?
The quotient module $R/Ra$ is generated by the coset of $1\in R$, which has order $a$. So it can also be written in the form $Rx$ where $x$ is an element of order $a$.