nLab submodule




Thoughout let RR be some ring. Write RRMod for the category of module over RR. Write URModU R Mod \to Set for the forgetful functor that sends a module to its underlying set.


For NRModN \in R Mod a module, a submodule of NN is a subset of U(N)U(N) which

  1. is a subgroup of the underlying group (closed under the addition in NN);

  2. is preserved by the RR-action.

Equivalently this means:


A submodule of NRModN \in R Mod is a module homomorphism i:KNi : K \to N whose underlying map U(i)U(i) of sets is an injection.

And since the injections in RRMod are precisely the monomorphisms, this means that equivalently


A submodule of NRModN \in R Mod is a monomorphism i:KNi : K \hookrightarrow N in RRMod. Hence a submodule is a subobject in RRMod.


Given a submodule KNK \hookrightarrow N, the quotient module NK\frac{N}{K} is the quotient group of the underlying abelian groups.



For N=RN = R regarded as a module over itself, a submodule is precisely an ideal of RR.


For f:N 1N 2f : N_1 \to N_2 a homomorphism of modules,

  1. the kernel ker(f)N 1ker(f) \hookrightarrow N_1 is a submodule of N 1N_1,

  2. the image im(f)N 2im(f) \hookrightarrow N_2 is a submodule of N 2N_2.


In example quotient module of N 2N_2 by the image is the cokernel of ff

coker(f)N 2im(f). coker(f) \simeq \frac{N_2}{im(f)} \,.


Of free modules

Let RR be a ring.


Assuming the axiom of choice, the following are equivalent

  1. every submodule of a free module over RR is itself free;

  2. every ideal in RR is a free RR-module;

  3. RR is a principal ideal domain.

A proof is in (Rotman, pages 650-651).


For instance

  • Rotman Advanced Modern Algebra

Last revised on November 21, 2022 at 07:03:08. See the history of this page for a list of all contributions to it.