3

I am trying to solve the following qualifying exam problem: “Give an example of a module over a PID that is not isomorphic to a direct sum of cyclic modules. Justify your example”. (Carnegie Mellon, 2022 Algebra Qual Q1)

My attempt: By the Structure Theorem of Finitely Generated Modules over PIDs, if $M$ is a finitely generated module over a PID $R$, then $$M \cong R^n \oplus R/p_1^{n_1} \oplus \dotsb \oplus R / p_k^{n_k} $$ where $n_1, \dotsc, n_k \ge 0$ and $p_1, \dotsc, p_k$ are primes in $R$. Thus, we are looking for a module $M$ over a PID $R$, which is not finitely generated. Can I just pick any module which is not finitely generated? Or should I be more careful about the choice?

Thanks in advance.

Jendrik Stelzner
  • 17,081
ALNS
  • 269

2 Answers2

5

Try $\mathbb{Q}$ over $\mathbb{Z}$, which is a PID. Note that any two cyclic submodules of $\mathbb{Q}$ intersects nontrivially.

Kiyoshi
  • 96
  • 3
    In my mind this is more of a comment than an answer - proving that $\mathbb Q$ is not a direct sum of cyclic modules (i.e. cyclic Abelian groups) is not hard, but the OP may not be able to see it immediately. Luckily, there is a proof here on MSE. –  Dec 31 '22 at 10:00
4

Actually if $R$ is any PID that is not a field, then its fraction field $F(R)$ is an example. If $F(R)$ is a direct sum of cyclic $R$-modules, since it's torsion-free, it must be free, i.e. $F(R)\simeq \oplus_{i\in I}R$ where $I\not=\emptyset$. Then we pick a nonzero noninvertible element $r\in R$, and we have $F(R)\otimes R/(r)=\{0\}$ while $(\oplus_{i\in I}R)\otimes R/(r)\simeq\oplus_{i\in I}R/(r)\not=\{0\}$.

From another point of view, $F(R)$ is injective, and hence cannot be projective or free.

Another famous result is that $\prod_{i=1}^\infty\mathbb Z$ is not a free $\mathbb Z$-module, but this is much harder.

Just a user
  • 22,048