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.

Let M be a finitely generated R-module ,where R is a PID and N a submodule.Is there an epimorphism from M to N?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

Yes, but the epimorphism might not doing anything nice to N ≤ M. Just use the invariants. What do the invariants of submodules look like? What do the invariants of quotient modules look like?

For instance R=Z, M=ZxZxZ/4Z has invariants (0,0,4) and its submodules have invariants (0,0,4), (0,0,2), (0,0), (0,4), (0,2), (0), (4), (2), and (). Its quotient modules have many invariants, but amongst them are all of the submodules' invariants.

You could also handle the free and torsion parts separately to get a cleaner proof. The submodules of the free summand are easy to understand, being free of at most the same rank, and clearly each is isomorphic to a quotient module. The submodule lattice and quotient module lattice of the torsion summand are isomorphic, so you can do better than just "some random epimorphism".

Invariants of submodules of (bounded) torsion modules: If R is a PID and M is a torsion module, then M is the direct sum of its submodules Mp = { m in M : there is an n such that pn⋅m = 0 } where p varies over the set of (nonzero) primes. If N is a submodule of M, then Np = N∩Mp by definition. To study submodules of torsion modules, it suffices to study the submodules of their p-parts, Mp.

If Mp is finitely generated, then there is a largest n needed to get pn⋅m = 0. In other words, we are looking at modules M such that pnM = 0, for some prime p in the PID R (or any noetherian, commutative ring where p is a principal maximal ideal). Such an M decomposes into a direct sum M = (R/pR)(r1) ⊕ (R/p2R)(r2) ⊕⋯⊕ (R/pnR)(rn). Call such a module (p; r1, r2, …, rn).

Details: Two such modules are isomorphic iff after removing any "trailing 0s" (that is, with rn ≠ 0) the labels are identical. Any such labeling represents a bounded module as long as the ri are all cardinal numbers, and the module is finitely generated iff the ri are all finite.

The isomorphism classes of its submodules are precisely the (p; s1, s2, …, sn) with 0 ≤ si + si+1 + … sn ≤ ri + ri+1 + … rn.

For example Z/9Z × Z/3Z × Z/3Z has label (3;2,1) and its submodules have labels (3;0,0), (3;0,1), (3;1,0), (3;1,1), (3;2,0), (3;2,1), and (3;3,0). The submodule Z/9Z × Z/3Z × 1 for instance has label (3;1,1), the submodule Z/9Z × 1 × 1 has label (3;0,1), and the submodule 3Z/9Z × Z/3Z × Z/3Z has label (3;3,0).

share|improve this answer
This was indeed easy!Thank you for your time –  t.k Jan 5 '11 at 19:06
No problem. Sometimes it is handy to know the lattice duality thingy, but sometimes it is just more complication: Fix an isomorphism f between Hom(M,K/R) and M, where R is a PID and K is its field of fractions and M is a finitely generated torsion module. Then A submodule N ≤ M creates a quotient module Hom(N,K/R) of Hom(M,K/R) and you apply f to get a quotient module of M. –  Jack Schmidt Jan 5 '11 at 19:10
add comment

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.