Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $p$ is irreducible and $M$ is a tosion module over a PID $R$ that can be written as a direct sum of cyclic submodules with annihilators of the form $p^{a_1} | \cdots | p^{a_s}$ and $p^{a_i}|p^{a_i+1}$. Let now $N$ be a submodule of $M$. How can i prove that $N$ can be written a direct sum of cyclic modules with annihilators of the form $p^{b_1} | \cdots | p^{b_t}, t\leq s$ and $\ p^{b_i}| p^ {a_(s-t+i)}$?

I've already shown that $t\leq s$ considering the epimorphism from a free module to $M$ and from its submodule to $N$.

share|improve this question
Please do not post commands to the group; state your questions as questions, explaining why you are considering it and where you are stuck or what your thoughts are. If the problem is homework, please tag it as such as well. –  Arturo Magidin Jan 12 '11 at 2:43
You were wright about the commands and i apologise.That's why I changed the question. –  t.k Jan 12 '11 at 8:38
@t.spero: And please use appropriate mark-up. Use \leq, not <=, for instance. –  Arturo Magidin Jan 12 '11 at 17:15
@A.Magidin:Where can i find a guide with the mark-up? –  t.k Jan 13 '11 at 17:33
@t.spero: for the pinging to work, the first few characters have to match the user name; so @A.Magidin did not notify me of your comment. As to your question, for HTML markup there is a link to the right of the posting window; for mathematics mark-up, it is basic LaTeX, so any tutorial on LaTeX will do. –  Arturo Magidin Jan 18 '11 at 19:26

4 Answers 4

Here's an ugly argument that I know of (which works for abelian groups, and so I assume it works over arbitrary PID). You need to make three observations:

  1. Maximal cyclic $p$-submodules of a $p$-module $M$ are the direct summands of $M$.
  2. Every cyclic $p$-submodule of $M$ is contained in a maximal cyclic $p$-submodule.
  3. If $N_1\subset M_1$ and $N_2\subset M_2$ are cyclic $p$-submodules of $M$, then $M_1$ and $M_2$ are disjoint if and only if $N_1$ and $N_2$ (this follows from observing that the submodules of a cyclic $p$-module are totally ordered under inclusion, and considering where $M_1\cap M_2$ is relative to $N_1$ and $N_2$ as a submodule of $M_1$ or $M_2$; you may have to use the fact that cyclic $p$-modules are indecomposable).

Once you have proven those, take a direct sum decomposition of $N$ into cylic $p$-submodules $N_1\oplus N_2\oplus\dots\oplus N_i$ (which exists by structure theorem of finitely generated modules over a PID), and use 1., 2. and 3. to obtain a decomposition $M=M_1\oplus\dots M_n\oplus M'$, where $M_i$ is the maximal cyclic $p$-submodule that contains the cyclic $p$-submodule $N_i$.

*by $p$-module I mean a module annihilated by some power of the prime $p$ in the PID $R$ (it's the primality of $p$ that matters, not irreducibility, for the total ordering of submodules, though the two coincide over PIDs, which is where you need to be anyway for the structure theorem to work)

share|improve this answer

See Lemma 9 here.

share|improve this answer
I don't see how this would work –  t.k Jan 12 '11 at 7:04
did you misread the question (thinking it was about the structure theorem) or is there actually a nice proof of the decomposition of submodules of torsion modules that follows from Lemma 9? –  Vladimir Sotirov Jan 12 '11 at 23:43

just look at the size of the subgroups annihilated by p^k for various k.

share|improve this answer
I don't think this works because torsion groups are finite in contrast with modules in general. –  t.k Jan 14 '11 at 15:00
by "size" I meant the number of summands in its decomposition. E.g. a submodule of R/p^2 + R/p^6 cannot decompose as R/p^3 + R/p^4, because the submodule annihilated by p^3 has too many summands. and you've already proved that part I think. Does this make sense? If not, my apologies. –  roy smith Jan 16 '11 at 21:46
or maybe in my comment two above one should look at the number of components of the image of the map p^2. –  roy smith Jan 17 '11 at 5:37
was that clear? you say you have proved the decomposition of a submodule cannot have more summands than the larger module. but then the image of multiplication by p^2 on the example submodule above, would have 2 summands, while the larger image of that map on the original module would have only one. this example seems to generalize. –  roy smith Jan 17 '11 at 22:48
Your idea reminds me of a proof of the second decomposition theorem of f.g modules.I think it works(my problem with your answer was the "size" word because discussing it with colleagues came up problems when we tried to generalize from groups where size matters). –  t.k Jan 18 '11 at 8:28

Here's a proof :

The result can be easily shown by strong induction to $\sum_{i=1}^{s}{a_i}$ .Suppose that the result is true if the sum above is less or equal to k.Suppose now that the sum is equal to k+1.Then for the induction step we use the invariant factors of pM for which the sum is less than k.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.