It is well-known that if $M$ is a finitely generated module over a PID $R$, then $M\cong R/R{p_1}^{n_1}\oplus \ldots \oplus R/R{p_k}^{n_k}\oplus F$ for some prime element $p_i$ and for some positive integers $n_i$ $(1\leq i\leq k)$ and free $R$-module $F$ finite rank. What is the form of a submodule $N$ of $M$ with respect to this characterization?


A submodule is again a finitely generated module over a PID, hence it is of the form

$N \cong R/q_1^{m_1} \oplus \dotsb \oplus R/q_s^{m_s} \oplus F^\prime$

with $m_i \geq 1$ and $F^\prime$ free.

By tensoring with the quotient field, it is clear that the rank of $F^\prime$ is at most the rank of $F$.

By localizing the injection $N \hookrightarrow M$ at the prime $(q_j)$, we obtain that each $q_j$ occurs within the $p_j$. If not, the right hand side would have killed all torsion, while the left hand side still had torsion.

Thus we can write

$$N \cong R/p_1^{m_1} \oplus \dotsb \oplus R/p_k^{m_k} \oplus F^\prime$$

with $m_i \geq 0$.

By localizing at $(p_i)$ and then passing to the torsion part, we get an injection $R/p_i^{m_i} \hookrightarrow R/p_i^{n_i}$.

In particular $1 \in R/p_i^{m_i}$ is annihilated by $p_i^{n_i}$, hence $p_i^{n_i} \in (p_i^{m_i})$ or equivalently $m_i \leq n_i$.

Summarizing, we get the (intuitively obvious) result, that $N$ is of the form

$$N \cong R/p_1^{m_1} \oplus \dotsb \oplus R/p_k^{m_k} \oplus F^\prime$$

with $0 \leq m_i \leq n_i$ and $\operatorname{rank} F^\prime \leq \operatorname{rank}F$.

  • 1
    $\begingroup$ It's not quite so simple, because the $p_i$ might not all be distinct, so that localizing at one of them does not isolate a single torsion summand of $M$. You have to look at the dimension of $p_i^nN/p_i^{n+1}N$ for each $n$ to entirely determine what the $m_i$ can be. $\endgroup$ – Eric Wofsey Nov 17 '16 at 10:21

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