Here, we will prove that the process of decomposing is unique, given that *M* is noetherian and artinian.

Again, *R* is a ring, possibly non-commutative.

Definition. Adecompositionof an R-module M is an expression for non-zero modulesAn R-module M is said to be

indecomposableif no such expression exists for k≥2 (or equivalently, for k=2).

A simple module is clearly indecomposable, but the converse is not true.

For example, let be the ring of upper-triangular 2 × 2 real matrices. Consider the module *M* of column vectors **R**^{2}, where *R* acts by matrix multiplication. Then *M* is not simple since it has a submodule spanned by (1, 0). But *M* is indecomposable since any submodule *N* ⊂ *M* which contains an (*x*, *y*) for *y*≠0 must be the whole *M*.

We have a criterion for a module to be indecomposable:

Proposition. Let M be an artinian and noetherian R-module. Then M is indecomposable if and only if is a local ring.

**Proof**

⇐ : if *M* = *M*_{1} ⊕ *M*_{2} is a decomposition of *M*, let π_{1}, π_{2} : *M* → *M* be the projection onto *M*_{1}, *M*_{2} respectively. Then π_{1}, π_{2} are not units but π_{1}+π_{2} is.

⇒ : recall that for an artinian and noetherian *M*, a module map *f* : *M* → *M* is bijective iff it is surjective or injective. Now suppose *f*, *g* : *M* → *M* are module maps such that *f*+*g* is a unit and *g* is not. We need to show that *f* is a unit.

Write (*f*+*g*)*h* = 1 for some *h* : *M* → *M*, so *fh* = 1-*gh*. Now *k* := *gh* is not an isomorphism since *g* is not surjective. Since *M* is artinian and noetherian, eventually

for some large *n*. We claim that

- To show : for any
*x*∈*M*, we have so we can write But then ; clearly the second term lies in and the first term gives so it lies in

- To show : if lies in then we have which gives and so

Since *M* is indecomposable either or The former case says *k* is injective and hence an isomorphism, so we have the latter case *k ^{n}*=0. Thus

*fh*= 1-

*k*is invertible (since 1 + nilpotent = unit), so

*f*is right-invertible. By symmetry, it is also left-invertible. ♦

## The Krull-Schmidt Theorem

Before stating and proving our main theorem, here is a useful criterion for splitting a module as a direct sum:

Splitting Lemma. If satisfy then N splits as Furthermore f is injective, thus M is a direct summand of N.

**Note**

To remember this, imagine *N* = *M*⊕*P* with *f* : *M* → *M*⊕*P* and *g* : *M*⊕*P* → *M* given by the inclusion and projection maps.

**Proof**

*f* is injective since *gf* is. For the first statement:

*M*= im(*f*) + ker(*g*) : any*x*∈*M*can be written as*x*=*fg*(*x*) + (*x*–*fg*(*x*)); the first term lies in im(*f*); the second term gives*g*(*x*–*fg*(*x*)) =*g*(*x*) –*gfg*(*x*) = 0.

- im(
*f*) ∩ ker(*g*) = 0: if*f*(*y*) lies in ker(*g*), then*gf*(*y*) = 0 so*y*= 0. ♦

Finally the main theorem we would like to prove:

Krull-Schmidt Theorem. Let M be an artinian and noetherian module. If we decompose:then k=l and there is a permutation σ of {1, 2, …, k} such that for all i=1,…,k.

**Proof**

We shall prove that *U*_{1} ≅ *V _{j}* for some

*j*, such that after which we can apply induction on

*k*.

Let be the projection onto *U*_{1} and be the projection onto *V _{i}*. This gives and thus Since im(

*p*) ⊆

*U*

_{1}this restricts to a map

Since *U*_{1} is indecomposable, End(*U*_{1}) is local; and since , one of the α* _{j}* is invertible, say α

_{1}. Let be the inverse of , so

where (*)

From the splitting lemma, *U*_{1 }is a direct summand of *V*_{1}; since *V*_{1} is indecomposable we must have *U*_{1}≅*V*_{1}. To prove that we write (*) as

where

Again, splitting lemma tells us The second term is isomorphic to *V*_{1} and the first is just ker(*p*), which is as desired. ♦

**Note**

Because *M* is artinian and noetherian it is always possible to decompose *M* as a direct sum of *finitely many* indecomposable modules. Simply keep decomposing the terms until we’re left with indecomposable modules: since *M* is of finite length, the process must terminate. Krull-Schmidt theorem says that the resulting terms are unique to isomorphism and permutation.

**Example**

Let *R* be the ring of upper-triangular 2 × 2 matrices with real entries. Here are two different ways of decomposing *R* as a direct sum of indecomposable left ideals:

Regardless of the decomposition we pick, the second term must be there (prove this!).