Homomorphisms, Quotients, and the Isomorphism Theorems

R-linear maps and the Hom module

A module homomorphism (or R-linear map) f : M → N is a group homomorphism of the underlying abelian groups that also commutes with the action: f(rm) = r f(m). When R = k this is exactly a linear transformation; when R = Z it is exactly a group homomorphism — once again the module notion swallows both. The kernel ker f = { m : f(m) = 0 } is a submodule of M, the image im f a submodule of N.

Something subtle and useful: the set Hom_R(M, N) of all R-linear maps is itself an abelian group under pointwise addition, and when R is commutative it is an R-module via (rf)(m) = r·f(m). This self-referential richness — Hom of modules being a module — is what makes the homological machinery of the later guides possible. The whole setup organizes into a category R-Mod whose objects are modules and arrows are R-linear maps.

Quotient modules and the four theorems

Because submodules are automatically “normal” (the group is abelian, so no normality worry), every submodule N ⊆ M yields a quotient module M/N: cosets m + N with action r(m+N) = rm + N, well-defined precisely because N is a submodule. The quotient map M → M/N is R-linear with kernel N. From here the isomorphism theorems are forced, exactly mirroring the group and ring cases.

1. First. For f : M → N, M/ker f ≅ im f. The quotient by the kernel recovers the image — the universal “factor through the kernel” statement.
2. Second (diamond). For submodules A, B: (A+B)/B ≅ A/(A∩B).
3. Third. For N ⊆ L ⊆ M: (M/N)/(L/N) ≅ M/L — “cancel the N.”
4. Fourth (correspondence). Submodules of M/N correspond bijectively, order-preservingly, with submodules of M containing N.

Worked instance of the First Isomorphism Theorem, R = Z.

Define f : Z -> Z/6Z by f(n) = n mod 6. This is Z-linear and onto.
  ker f = { n : n = 0 mod 6 } = 6Z
  im f  = Z/6Z
First theorem:  Z / 6Z  ≅  Z/6Z.   (a tautology here, but it CHECKS)

Now the Third theorem with N = 6Z, L = 2Z, M = Z (so 6Z ⊆ 2Z ⊆ Z):
  M/N = Z/6Z   has 6 elements {0,1,2,3,4,5}
  L/N = 2Z/6Z = {0,2,4}   the even classes, a submodule of order 3
  (M/N)/(L/N) = (Z/6Z)/{0,2,4}   has 6/3 = 2 elements
  M/L = Z/2Z   also has 2 elements
Indeed (Z/6Z)/(2Z/6Z) ≅ Z/2Z.   The N's cancel.

Simple modules: the atoms

A nonzero module is simple if its only submodules are 0 and itself — the module analogue of a simple group. By the correspondence theorem, simple modules are exactly the quotients R/m for a maximal ideal m: cyclic, and with no proper nonzero submodule because m is maximal. These are the indivisible building blocks; Guide 4 stacks them into composition series.