We have seen that for a semisimple algebra, any non-zero module is a direct sum of simple modules (see Theorem 4.11). We investigate now how this generalizes when we consider finite-dimensional modules. If the algebra is not semisimple, one needs to consider indecomposable modules instead of just simple modules, and then one might hope that any finite-dimensional module is a direct sum of indecomposable modules. We will show that this is indeed the case. In addition, we will show that a direct sum decomposition into indecomposable summands is essentially unique; this is known as the Krull–Schmidt Theorem.

In Chap. 3 we have studied simple modules, which are building blocks for arbitrary modules. They might be thought of as analogues of ‘elementary particles’, and then indecomposable modules could be viewed as analogues of ‘molecules’.

Throughout this chapter, *A* is a *K*-algebra where *K* is a field.

## 7.1 Indecomposable Modules

In this section we define indecomposable modules and discuss several examples. In addition, we will show that every finite-dimensional module is a direct sum of indecomposable modules.

### Definition 7.1.

Let *A* be a *K*-algebra, and assume *M* is a non-zero *A*-module. Then *M* is called *indecomposable* if it cannot be written as a direct sum *M* = *U* ⊕ *V* for non-zero submodules *U* and *V* . Otherwise, *M* is called *decomposable*.

### Remark 7.2.

- (1)
Every simple module is indecomposable.

- (2)Consider the algebra
*A*=*K*[*X*]∕(*X*^{t}) for some*t*≥ 2, recall that the*A*-modules are of the form*V*_{α}with*α*the linear map of the underlying vector space*V*given by the action of the coset of*X*, and note that*α*^{t}= 0. Let*V*_{α}be the 2-dimensional*A*-module where*α*has matrixwith respect to some basis. It has a unique one-dimensional submodule (spanned by the first basis vector). So it is not simple, and it is indecomposable, since otherwise it would be a direct sum of two 1-dimensional submodules. - (3)
Let

*A*=*K*[*X*]∕(*X*^{t}) where*t*≥ 2, and let*M*=*A*as an*A*-module. By the submodule correspondence, every non-zero submodule of*M*is of the form (*g*)∕(*X*^{t}) where*g*divides*X*^{t}but*g*is not a scalar multiple of*X*^{t}. That is, we can take*g*=*X*^{r}for*r*<*t*. We see that any such submodule contains the element*X*^{t−1}+ (*X*^{t}). This means that any two non-zero submodules of*M*have a non-zero intersection. Therefore*M*must be indecomposable. - (4)
Let

*A*be a semisimple*K*-algebra. Then an*A*-module is simple if and only if it is indecomposable. Indeed, by (1) we know that a simple module is indecomposable. For the converse, let*M*be an indecomposable*A*-module and let*U*⊆*M*be a non-zero submodule; we must show that*U*=*M*. Since*A*is a semisimple algebra, the module*M*is semisimple (see Theorem 4.11). So by Theorem 4.3, the submodule*U*has a complement, that is,*M*=*U*⊕*C*for some*A*-submodule*C*of*M*. But*M*is indecomposable, and*U*≠ 0, so*C*= 0 and then*U*=*M*.

Suppose *A* is a *K*-algebra and *M* an *A*-module such that *M* = *U* ⊕ *V* with *U*, *V* non-zero submodules of *M*. Then in particular, this is a direct sum of *K*-vector spaces. In linear algebra, having a direct sum decomposition *M* = *U* ⊕ *V* of a non-zero vector space *M* is the same as specifying a projection, *ε* say, which maps *V* to zero, and is the identity on *U*. If *U* and *V* are *A*-submodules of *M* then *ε* is an *A*-module homomorphism, for example by observing that *ε* = *ι* ∘ *π* where *π* : *M* → *U* is the canonical surjection, and *ι* : *U* → *M* is the inclusion homomorphism (see Example 2.22). Then *U* = *ε*(*M*) and *V* = (id_{M} − *ε*)(*M*), and *ε*
^{2} = *ε*, and (id_{M} − *ε*)^{2} = id_{M} − *ε*.

Recall that an idempotent of an algebra is an element *ε* in this algebra such that *ε*
^{2} = *ε*. We see that from the above direct sum decomposition of *M* as an *A*-module we get an idempotent *ε* in the endomorphism algebra End_{A}(*M*).

### Lemma 7.3.

*Let A be a K-algebra, and let M be a non-zero A-module. Then M is indecomposable if and only if the endomorphism algebra*
End_{A}(*M*) *does not contain any idempotents except* 0 *and* id_{M}.

### Proof.

*M*is indecomposable. Suppose

*ε*∈End

_{A}(

*M*) is an idempotent, then since id

_{M}=

*ε*+(id

_{M}−

*ε*) we have

*M*=

*ε*(

*M*)+(id

_{M}−

*ε*)(

*M*). Moreover, this sum is direct: if

*x*∈

*ε*(

*M*) ∩ (id

_{M}−

*ε*)(

*M*) then

*ε*(

*m*) =

*x*= (id

_{M}−

*ε*)(

*n*) with

*m*,

*n*∈

*M*, and then

*M*=

*ε*(

*M*) ⊕ (id

_{M}−

*ε*)(

*M*) is a direct sum of

*A*-submodules. Since

*M*is indecomposable,

*ε*(

*M*) = 0 or (id

_{M}−

*ε*)(

*M*) = 0. That is,

*ε*= 0 or else id

_{M}−

*ε*= 0, which means

*ε*= id

_{M}.

For the converse, if *M* = *U* ⊕ *V* , where *U* and *V* are submodules of *M*, then as we have seen above, the projection *ε* : *M* → *M* with *ε*(*u* + *v*) = *u* for *u* ∈ *U*, *v* ∈ *V* is an idempotent in End_{A}(*M*). By assumption, *ε* is zero or the identity, which means that *U* = 0 or *V* = 0. Hence *M* is indecomposable.□

### Example 7.4.

*A*:=

*KQ*of the Kronecker quiver as in Example 1.13,

- (1)We consider the
*A*-submodule*M*:=*Ae*_{1}= span{*e*_{1},*a*,*b*} of*A*, and*U*:= span{*a*,*b*}. Each element in the basis {*e*_{1},*e*_{2},*a*,*b*} of*A*acts on*U*by scalars, and hence*U*and every subspace of*U*is an*A*-submodule, and*U*is the direct sum of non-zero*A*-submodulesHowever,*M*is indecomposable. We will prove this using the criterion of Lemma 7.3. Let*ε*:*M*→*M*be an*A*-module homomorphism with*ε*^{2}=*ε*, then we must show that*ε*= id_{M}or*ε*= 0. We have*ε*(*e*_{1}*M*) =*e*_{1}*ε*(*M*) ⊆*e*_{1}*M*, but*e*_{1}*M*is spanned by*e*_{1}, so*ε*(*e*_{1}) =*λe*_{1}for some scalar*λ*∈*K*. Next, note*a*=*ae*_{1}and thereforeSimilarly, since*b*=*be*_{1}we have*ε*(*b*) =*λb*. We have proved that*ε*=*λ*⋅id_{M}. Now,*ε*^{2}=*ε*and therefore*λ*^{2}=*λ*and hence*λ*= 0 or*λ*= 1. That is,*ε*is the zero map, or the identity map. - (2)Let
*N*be the*A*-module with basis where*e*_{1}*N*has basis and*e*_{2}*N*has basis {*v*_{2}} and where the action of*a*and*b*is defined byWe see that*U*:= span{*v*_{2}} is an*A*-submodule of*N*. The factor module*N*∕*U*has basis consisting of the cosets of , and from the definition, each element in the basis {*e*_{1},*e*_{2},*a*,*b*} of*A*acts by scalar multiplication on*N*∕*U*. As before, this implies that*N*∕*U*is a direct sum of two 1-dimensional modules. On the other hand, we claim that*N*is indecomposable; to show this we want to use Lemma 7.3, as in (1). So let*ε*∈End_{A}(*N*), then*ε*(*v*_{2}) =*ε*(*e*_{2}*v*_{2}) =*e*_{2}*ε*(*v*_{2}) ∈*e*_{2}*N*and hence*ε*(*v*_{2}) =*λv*_{2}, for some*λ*∈*K*. Moreover,*ε*(*v*_{1}) =*ε*(*e*_{1}*v*_{1}) =*e*_{1}*ε*(*v*_{1}) ∈*e*_{1}*N*, so for some*μ*,*ρ*∈*K*. Using that*bv*_{1}= 0 and this implies that*ρ*= 0. Then*λv*_{2}=*ε*(*v*_{2}) =*ε*(*av*_{1}) =*aε*(*v*_{1}) =*μv*_{2}, hence*λ*=*μ*and*ε*(*v*_{1}) =*λv*_{1}. Similarly, one shows , that is, we get*ε*=*λ*⋅id_{N}. If*ε*^{2}=*ε*then*ε*= 0 or*ε*= id_{N}.

We will now show that every non-zero finite-dimensional module can be written as a direct sum of indecomposable modules.

### Theorem 7.5.

*Let A be a K-algebra, and let M be a non-zero finite-dimensional A-module. Then M can be expressed as a direct sum of finitely many indecomposable A-submodules.*

### Proof.

We use induction on the vector space dimension dim_{K}
*M*. If dim_{K}
*M* = 1, then *M* is a simple *A*-module, hence is an indecomposable *A*-module, and we are done. So let dim_{K}
*M* > 1. If *M* is indecomposable, there is nothing to do. Otherwise, *M* = *U* ⊕ *V* with non-zero *A*-submodules *U* and *V* of *M*. Then both *U* and *V* have strictly smaller dimension than *M*. Hence by the inductive hypothesis, each of *U* and *V* can be written as a direct sum of finitely many indecomposable *A*-submodules. Since *M* = *U* ⊕ *V* , it follows that *M* can be expressed as a direct sum of finitely many indecomposable *A*-submodules.□

### Remark 7.6.

There is a more general version of Theorem 7.5. Recall from Sect. 3.3 that an *A*-module *M* is said to be of finite length if *M* has a composition series; in this case, the length *ℓ*(*M*) of *M* is defined as the length of a composition series of *M* (which is uniquely determined by the Jordan–Hölder theorem). If we replace ‘dimension’ in the above theorem and its proof by ‘length’ then everything works the same, noting that proper submodules of a module of finite length have strictly smaller lengths, by Proposition 3.17. We get therefore that any non-zero module of finite length can be expressed as a direct sum of finitely many indecomposable submodules.

There are many modules which cannot be expressed as a finite direct sum of indecomposable modules. For example, let , and let be the set of real numbers. Then is a vector space over , hence is a -module. The indecomposable -modules are the 1-dimensional -vector spaces. Since has infinite dimension over , it cannot be a finite direct sum of indecomposable -modules. This shows that the condition that the module is finite-dimensional (or has finite length) in Theorem 7.5 cannot be removed.

## 7.2 Fitting’s Lemma and Local Algebras

We would like to have criteria which tell us when a given module is indecomposable. Obviously Definition 7.1 is not so helpful; we would need to inspect all submodules of a given module. One criterion is Lemma 7.3; in this section we will look for further information from linear algebra.

Given a linear transformation of a finite-dimensional vector space, one gets a direct sum decomposition of the vector space, in terms of the kernel and the image of some power of the linear transformation.

### Lemma 7.7 (Fitting’s Lemma I).

*Let K be a field. Assume V is a finite-dimensional K-vector space, and θ*:

*V*→

*V is a linear transformation. Then there is some n*≥ 1

*such that the following hold:*

- (i)
*For all k*≥ 0*we have**and*im(*θ*^{n}) = im(*θ*^{n+k}). - (ii)
.

### Proof.

This is elementary linear algebra, but since it is important, we give the proof.

*V*is finite-dimensional, the ascending chain (7.1) cannot contain infinitely many strict inequalities, so there exists some

*n*

_{1}≥ 1 such that for all

*k*≥ 0. Similarly, for the descending chain (7.2) there is some

*n*

_{2}≥ 1 such that for all

*k*≥ 0. Setting

*n*as the maximum of

*n*

_{1}and

*n*

_{2}proves (i).

*θ*

^{n}(

*x*) = 0 and

*x*=

*θ*

^{n}(

*y*) for some

*y*∈

*V*. We substitute, and then we have that 0 =

*θ*

^{n}(

*x*) =

*θ*

^{2n}(

*y*). This implies by part (i), and thus

*x*=

*θ*

^{n}(

*y*) = 0. Hence . Now, by the rank-nullity theorem we have

*V*since it is a subspace whose dimension is equal to dim

_{K}

*V*.□

### Corollary 7.8.

*Let A be a K-algebra. Assume M is a finite-dimensional A-module, and θ *:* M *→* M is an A-module homomorphism. Then there is some n *≥ 1 *such that*
*, a direct sum of A-submodules of M.*

### Proof.

This follows directly from Lemma 7.7. Indeed, *θ* is in particular a linear transformation. The map *θ*
^{n} is an *A*-module homomorphism and therefore its kernel and its image are *A*-submodules of *M*.□

### Corollary 7.9 (Fitting’s Lemma II).

*Let A be a K-algebra and M a non-zero finite-dimensional A-module. Then the following statements are equivalent:*

- (i)
*M is an indecomposable A-module.* - (ii)
*Every homomorphism θ*∈End_{A}(*M*)*is either an isomorphism, or is nilpotent.*

### Proof.

We first assume that statement (i) holds. By Corollary 7.8, for every *θ* ∈End_{A}(*M*) we have *M* = ker(*θ*
^{n}) ⊕im(*θ*
^{n}) as *A*-modules, for some *n* ≥ 1. But *M* is indecomposable by assumption, so we conclude that ker(*θ*
^{n}) = 0 or im(*θ*
^{n}) = 0. In the second case we have *θ*
^{n} = 0, that is, *θ* is nilpotent. In the first case, *θ*
^{n} and hence also *θ* are injective, and moreover, *M* = im(*θ*
^{n}), therefore *θ*
^{n} and hence *θ* are surjective. So *θ* is an isomorphism.

Conversely, suppose that (ii) holds. To show that *M* is indecomposable, we apply Lemma 7.3. So let *ε* be an endomorphism of *M* such that *ε*
^{2} = *ε*. By assumption, *ε* is either nilpotent, or is an isomorphism. In the first case *ε* = 0 since *ε* = *ε*
^{2} = … = *ε*
^{n} for all *n* ≥ 1. In the second case, im(*ε*) = *M* and *ε* is the identity on *M*: If *m* ∈ *M* then *m* = *ε*(*y*) for some *y* ∈ *M*, hence *ε*(*m*) = *ε*
^{2}(*y*) = *ε*(*y*) = *m*.□

### Remark 7.10.

*E*:= End

_{A}(

*M*) of an indecomposable

*A*-module

*M*has the following property:

*if*

*a*∈

*E*

*is not invertible, then*1

_{E}−

*a*∈

*E*

*is invertible*. In fact, if

*a*∈

*E*is not invertible then it is nilpotent, by Corollary 7.9, say

*a*

^{n}= 0. Then we have

_{E}−

*a*is invertible.

Note that in a non-commutative algebra one has to be slightly careful when speaking of invertible elements. More precisely, one should speak of elements which have a left inverse or a right inverse, respectively. If for some element both a left inverse and a right inverse exist, then they coincide (since invertible elements form a group).

Algebras (or more generally rings) with the property in Remark 7.10 appear naturally in many places in mathematics. We now study them in some more detail.

### Theorem 7.11.

*Assume A is any K-algebra, then the following are equivalent:*

- (i)
*The set N of elements x*∈*A which do not have a left inverse is a left ideal of A.* - (ii)
*For all a*∈*A, at least one of a and*1_{A}−*a has a left inverse in A.*

We observe the following: Let *N* be as in (i). If *x* ∈ *N* and *a* ∈ *A* then *ax* cannot have a left inverse in *A*, since otherwise *x* would have a left inverse. That is, we have *ax* ∈ *N*, so that *AN* ⊆ *N*.

### Proof.

*N*is a left ideal of

*A*. By the above observation, we only have to show that

*N*is an additive subgroup of

*A*. Clearly 0 ∈

*N*. Now let

*x*,

*y*∈

*N*, assume (for a contradiction) that

*x*−

*y*is not in

*N*. Then there is some

*a*∈

*A*such that

*a*(

*x*−

*y*) = 1

_{A}, so that

*ax*does not have a left inverse, and therefore, using (ii) we deduce that (−

*a*)

*y*has a left inverse. But then

*y*has a left inverse, and

*y*∉

*N*, a contradiction. We have now shown that

*N*is a left ideal of

*A*.

Now assume (i) holds, we prove that this implies (ii). Assume *a* ∈ *A* does not have a left inverse in *A*. We have to show that then 1_{A} − *a* has a left inverse in *A*. If this is false then both *a* and 1_{A} − *a* belong to *N*. By assumption (i), *N* is closed under addition, therefore 1_{A} ∈ *N*, which is not true. This contradiction shows that 1_{A} − *a* must belong to *N*.□

### Definition 7.12.

A *K*-algebra *A* is called a *local algebra* (or just *local*)
if it satisfies the equivalent conditions from Theorem 7.11.

### Exercise 7.1.

Let *A* be a local *K*-algebra. Show that the left ideal *N* in Theorem 7.11 is a maximal left ideal of *A*, and that it is the only maximal left ideal of *A*.

### Remark 7.13.

Let *A* be a local *K*-algebra. By Exercise 7.1 the left ideal *N* in Theorem 7.11 is then precisely the Jacobson radical as defined and studied in Sect. 4.3 (see Definition 4.21). In particular, if *A* is finite-dimensional then this unique maximal left ideal is even a two-sided ideal (see Theorem 4.23).

### Lemma 7.14.

- (a)
*Assume A is a local K-algebra. Then the only idempotents in A**are*0*and*1_{A}. - (b)
*Assume A is a finite-dimensional algebra. Then A is local if and only if the only idempotents in A are*0*and*1_{A}.

### Proof.

(a) Let *ε* ∈ *A* be an idempotent. If *ε* has no left inverse, then by Theorem 7.11 we know that 1_{A} − *ε* has a left inverse, say *a*(1_{A} − *ε*) = 1_{A} for some *a* ∈ *A*. Then it follows that *ε* = 1_{A}
*ε* = *a*(1_{A} − *ε*)*ε* = *aε* − *aε*
^{2} = 0. On the other hand, if *ε* has a left inverse, say *bε* = 1_{A} for some *b* ∈ *A*, then *ε* = 1_{A}
*ε* = *bε*
^{2} = *bε* = 1_{A}.

_{A}are the only idempotents in

*A*. We will verify condition (ii) of Theorem 7.11, that is, let

*a*∈

*A*, then we show that at least one of

*a*and 1

_{A}−

*a*has a left inverse in

*A*. Consider the map

*θ*:

*A*→

*A*defined by

*θ*(

*x*) :=

*xa*. This is an

*A*-module homomorphism if we view

*A*as a left

*A*-module. By Corollary 7.8 we have for some

*n*≥ 1. So we have a unique expression

*ε*

_{i}are idempotents. By assumption,

*ε*

_{1}= 0 or

*ε*

_{1}= 1

_{A}. Furthermore, by Lemma 5.1 we have

*A*=

*Aε*

_{1}⊕

*Aε*

_{2}. If

*ε*

_{1}= 0 then

*A*= im(

*θ*

^{n}) =

*Aa*

^{n}and then

*a*has a left inverse in

*A*(since 1

_{A}=

*ba*

^{n}for some

*b*∈

*A*). Otherwise,

*ε*

_{1}= 1

_{A}and then , that is,

*a*

^{n}= 0. A computation as in Remark 7.10 then shows that 1

_{A}−

*a*has a left inverse, namely 1

_{A}+

*a*+

*a*

^{2}+ … +

*a*

^{n−1}.□

We will now investigate some examples.

### Example 7.15.

- (1)
Let

*A*=*K*, the one-dimensional*K*-algebra. Then for every*a*∈*A*, at least one of*a*or 1_{A}−*a*has a left inverse in*A*and hence*A*is local, by Theorem 7.11 and Definition 7.12. The same argument works to show that every division algebra*A*=*D*over the field*K*(see Definition 1.7) is local. - (2)
Let

*A*=*M*_{n}(*K*) where*n*≥ 2. Let*a*:=*E*_{11}, then*a*and 1_{A}−*a*do not have a left inverse. Hence*A*is not local. (We have*M*_{1}(*K*) =*K*, which is local, by (1).) - (3)
Consider the factor algebra

*A*=*K*[*X*]∕(*f*) for a non-constant polynomial*f*∈*K*[*X*]. Then*A*is a local algebra if and only if*f*=*g*^{m}for some*m*≥ 1, where*g*∈*K*[*X*] is an irreducible polynomial. The proof of this is Exercise 7.6.

### Exercise 7.2.

Assume *A* = *KQ*, where *Q* is a quiver with no oriented cycles. Show that *A* is local if and only if *Q* consists just of one vertex.

For finite-dimensional modules, we can now characterize indecomposability in terms of local endomorphism algebras.

### Corollary 7.16 (Fitting’s Lemma III).

*Let A be a K-algebra and let M be a non-zero finite-dimensional A-module. Then M is an indecomposable A-module if and only if the endomorphism algebra* End_{A}(*M*)
*is a local algebra.*

### Proof.

Let *E* := End_{A}(*M*). Note that *E* is finite-dimensional (it is contained in the space of all *K*-linear maps *M* → *M*, which is finite-dimensional by elementary linear algebra). By Lemma 7.3, the module *M* is indecomposable if and only if the algebra *E* does not have idempotents other than 0 and id_{M}. By Lemma 7.14 this is true if and only if *E* is local.□

### Remark 7.17.

All three versions of Fitting’s Lemma have a slightly more general version, if one replaces ‘finite dimensional’ by ‘finite length’, see Remark 7.6.

The assumption that *M* is finite-dimensional, or has finite length, cannot be omitted. For instance, consider the polynomial algebra *K*[*X*] as a *K*[*X*]-module. Multiplication by *X* defines a *K*[*X*]-module homomorphism *θ* : *K*[*X*] → *K*[*X*]. For every we have ker(*θ*
^{n}) = 0 and im(*θ*
^{n}) = (*X*
^{n}), the ideal generated by *X*
^{n}. But *K*[*X*] ≠ ker(*θ*
^{n}) ⊕im(*θ*
^{n}). So Lemma 7.7 fails for *A*. Exercise 7.11 contains further illustrations.

## 7.3 The Krull–Schmidt Theorem

We have seen in Theorem 7.5 that every non-zero finite-dimensional module can be decomposed into a finite direct sum of indecomposable modules. The fundamental Krull–Schmidt Theorem states that such a decomposition is unique up to isomorphism and up to reordering indecomposable summands. This is one of the most important results in representation theory.

### Theorem 7.18 (Krull–Schmidt Theorem).

*Let A be a K-algebra, and let M be a non-zero finite-dimensional A-module. Suppose there are two direct sum decompositions*

*of M into indecomposable A-submodules M*
_{i} (1 ≤* i *≤* r) and N*
_{j} (1 ≤* j *≤* s). Then r *=* s, and there is a permutation σ such that M*
_{i}≅*N*
_{σ(i)} *for all i *= 1, …, *r.*

Before starting with the proof, we introduce some notation we will use for the canonical homomorphisms associated to a direct sum decomposition, similar to the notation used in Lemma 5.6. Let *μ*
_{i} : *M* → *M*
_{i} be the homomorphism defined by *μ*(*m*
_{1} + … + *m*
_{r}) = *m*
_{i}, and let *ι*
_{i} : *M*
_{i} → *M* be the inclusion map. Then and hence if *e*
_{i} := *ι*
_{i} ∘ *μ*
_{i} : *M* → *M* then *e*
_{i} is the projection with image *M*
_{i} and kernel the direct sum of all *M*
_{j} for *j* ≠ *i*. Then the *e*
_{i} are orthogonal idempotents in End_{A}(*M*) with .

Similarly let *ν*
_{t} : *M* → *N*
_{t} be the homomorphism defined by *ν*
_{t}(*n*
_{1} + … + *n*
_{s}) = *n*
_{t}, and let *κ*
_{t} : *N*
_{t} → *M* be the inclusion. Then *ν*
_{t} ∘ *κ*
_{t} is the identity map of *N*
_{t}, and if *f*
_{t} := *κ*
_{t} ∘ *ν*
_{t} : *M* → *M* then *f*
_{t} is the projection with image *N*
_{t} and kernel the direct sum of all *N*
_{j} for *j* ≠ *t*. In addition, the *f*
_{t} are orthogonal idempotents in End_{A}(*M*) whose sum is the identity id_{M}.

In the proof below we sometimes identify *M*
_{i} with *ι*
_{i}(*M*
_{i}) and *N*
_{t} with *κ*
_{t}(*N*
_{t}), to ease notation.

### Proof.

We use induction on *r*, the number of summands in the first decomposition. When *r* = 1 we have *M*
_{1} = *M*. This means that *M* is indecomposable, and we conclude that *s* = 1 and *N*
_{1} = *M* = *M*
_{1}. Assume now that *r* > 1.

*N*

_{1}and some

*M*

_{i}. We have

*N*

_{1}is indecomposable and finite-dimensional, so by Corollary 7.9, every endomorphism of

*N*

_{1}is either nilpotent, or is an isomorphism. Assume for a contradiction that each summand in the above sum (∗) is nilpotent, and hence does not have a left inverse. We have that End

_{A}(

*N*

_{1}) is a local algebra, by Corollary 7.16. Hence by Theorem 7.11, the set of elements with no left inverse is closed under addition, so it follows that the sum also has no left inverse. But the sum is the identity of

*N*

_{1}which has a left inverse, a contradiction.

*M*

_{i}and assume that

*ϕ*is an isomorphism

*N*

_{1}→

*M*

_{1}, where

*M*=

*M*

_{1}, and

*N*

_{1}=

*N*=

*N*

^{′}and

*j*=

*μ*

_{1}∘

*κ*

_{1}and

*π*=

*ν*

_{1}∘

*ι*

_{1}. We obtain

*M*

_{1}is indecomposable and

*μ*

_{1}∘

*κ*

_{1}is non-zero we have

*M*

_{1}= im(

*μ*

_{1}∘

*κ*

_{1}) and . Hence the map

*μ*

_{1}∘

*κ*

_{1}:

*N*

_{1}→

*M*

_{1}is surjective. It is also injective (since

*ϕ*=

*ν*

_{1}∘

*ι*

_{1}∘

*μ*

_{1}∘

*κ*

_{1}is injective). This shows that

*μ*

_{1}∘

*κ*

_{1}is an isomorphism

*N*

_{1}→

*M*

_{1}.

*A*-module isomorphism

*γ*:

*M*→

*M*such that

*γ*(

*N*

_{1}) =

*M*

_{1}and

*γ*(

*N*

_{j}) =

*N*

_{j}for 2 ≤

*j*≤

*s*. Define

*γ*:

*M*→

*M*is an isomorphism. It suffices to show that

*γ*is injective, by dimensions. Let

*γ*(

*x*) = 0 for some

*x*∈

*M*. Using that

*f*

_{1}is an idempotent we have

*f*

_{1}∘

*e*

_{1}∘

*f*

_{1}=

*κ*

_{1}∘

*ν*

_{1}∘

*ι*

_{1}∘

*μ*

_{1}∘

*κ*

_{1}∘

*ν*

_{1}=

*κ*

_{1}∘

*ϕ*∘

*ν*

_{1}with the isomorphism

*ϕ*:

*N*

_{1}→

*M*

_{1}from (1). Since

*κ*

_{1}and

*ϕ*are injective, it follows from (

*f*

_{1}∘

*e*

_{1}∘

*f*

_{1})(

*x*) = 0 that

*ν*

_{1}(

*x*) = 0. Then also

*f*

_{1}(

*x*) = (

*κ*

_{1}∘

*ν*

_{1})(

*x*) = 0 and this implies

*x*=

*γ*(

*x*) = 0, as desired.

We now show that *γ*(*N*
_{1}) = *M*
_{1} and *γ*(*N*
_{j}) = *N*
_{j} for 2 ≤ *j* ≤ *s*. From (1) we have the isomorphism *μ*
_{1} ∘ *κ*
_{1} : *N*
_{1} → *M*
_{1}, and this is viewed as a homomorphism *e*
_{1} ∘ *f*
_{1} : *M* → *M*, by *e*
_{1} ∘ *f*
_{1} = *ι*
_{1} ∘ (*μ*
_{1} ∘ *κ*
_{1}) ∘ *ν*
_{1}, noting that *ν*
_{1} is the identity on *N*
_{1} and *ι*
_{1} is the identity on *M*
_{1}. Furthermore, id_{M} − *f*
_{1} maps *N*
_{1} to zero, and in total we see that *γ*(*N*
_{1}) = *M*
_{1}.

Moreover, if *x* ∈ *N*
_{j} and *j* ≥ 2 then *x* = *f*
_{j}(*x*) and *f*
_{1} ∘ *f*
_{j} = 0 and it follows that *γ*(*x*) = *x*. This proves *γ*(*N*
_{j}) = *N*
_{j}.

*M*

^{′}:=

*M*

_{2}⊕… ⊕

*M*

_{r}. We have obtained an isomorphism

*ψ*:

*M*

^{′}→

*N*

_{2}⊕… ⊕

*N*

_{s}. Let , this is a submodule of

*M*

^{′}, and we have, again by Exercise 7.3, the direct sum decomposition

*r*− 1 =

*s*− 1 and there is a permutation

*σ*of {1, 2, 3, …,

*r*} with for

*i*≥ 2 (and

*σ*(1) = 1). This completes the proof.□

### EXERCISES

- 7.3.
Let

*A*be a*K*-algebra, and*M*a non-zero*A*-module. Assume*M*=*U*_{1}⊕… ⊕*U*_{r}is a direct sum of*A*-submodules, and assume that*γ*:*M*→*M*is an*A*-module isomorphism. Show that then*M*=*γ*(*M*) =*γ*(*U*_{1}) ⊕… ⊕*γ*(*U*_{r}). - 7.4.Let
*T*_{n}(*K*) be the*K*-algebra of upper triangular*n*×*n*-matrices. In the natural*T*_{n}(*K*)-module*K*^{n}we consider for 0 ≤*i*≤*n*the submoduleswhere*e*_{i}denotes the*i*-th standard basis vector. Recall from Exercise 2.14 that*V*_{0},*V*_{1}, …,*V*_{n}are the only*T*_{n}(*K*)-submodules of*K*^{n}, and that*V*_{i,j}:=*V*_{i}∕*V*_{j}(for 0 ≤*j*<*i*≤*n*) are pairwise non-isomorphic*T*_{n}(*K*)-modules.- (a)
Determine the endomorphism algebra for all 0 ≤

*j*<*i*≤*n*. - (b)
Deduce that each

*V*_{i,j}is an indecomposable*T*_{n}(*K*)-module.

- (a)
- 7.5.
Recall that for any

*K*-algebra*A*and every element*a*∈*A*the map*θ*_{a}:*A*→*A*,*b*↦*ba*, is an*A*-module homomorphism.We consider the algebra*A*=*T*_{n}(*K*) of upper triangular*n*×*n*-matrices. Determine for each of the following elements*a*∈*T*_{n}(*K*) the minimal such that and for all . Moreover, give explicitly the decomposition (which exists by the Fitting Lemma, see Corollary 7.7):- (i)
*a*=*E*_{11}; - (ii)
*a*=*E*_{12}+*E*_{23}+ … +*E*_{n−1,n}; - (iii)
*a*=*E*_{1n}+*E*_{2n}+ … +*E*_{nn}.

- (i)
- 7.6.
- (a)
Let

*A*=*K*[*X*]∕(*f*) with a non-constant polynomial*f*∈*K*[*X*]. Show that*A*is a local algebra if and only*f*=*g*^{m}for some irreducible polynomial*g*∈*K*[*X*] and some (up to multiplication by a non-zero scalar in*K*). - (b)Which of the following algebras are local?
- (i)
, where

*n*≥ 2; - (ii)
where

*p*is a prime number; - (iii)
*K*[*X*]∕(*X*^{3}− 6*X*^{2}+ 12*X*− 8).

- (i)

- (a)
- 7.7.Which of the following
*K*-algebras*A*are local?- (i)
*A*=*T*_{n}(*K*), the algebra of upper triangular*n*×*n*-matrices; - (ii)
*A*= {*a*= (*a*_{ij}) ∈*T*_{n}(*K*) |*a*_{ii}=*a*_{jj}for all*i*,*j*}.

- (i)
- 7.8.For a field
*K*let be the set of formal power series. On*K*[[*X*]] define the following operations:addition (∑

_{i}*λ*_{i}*X*^{i}) + (∑_{i}*μ*_{i}*X*^{i}) =∑_{i}(*λ*_{i}+*μ*_{i})*X*^{i},scalar multiplication

*λ*(∑_{i}*λ*_{i}*X*^{i}) =∑_{i}*λλ*_{i}*X*^{i},multiplication (∑

_{i}*λ*_{i}*X*^{i})(∑_{j}*μ*_{j}*X*^{j}) =∑_{k}(∑_{i+j=k}*λ*_{i}*μ*_{j})*X*^{k}.

- (a)
Verify that

*K*[[*X*]] with these operations becomes a commutative*K*-algebra. - (b)
Determine the invertible elements in

*K*[[*X*]]. - (c)
Show that

*K*[[*X*]] is a local algebra.

- 7.9.
Let

*K*be a field and*A*a*K*-algebra of finite length (as an*A*-module). Show that if*A*is a local algebra then*A*has only one simple module, up to isomorphism. - 7.10.For a field
*K*consider the path algebra*KQ*of the Kronecker quiverFor*λ*∈*K*we consider a 2-dimensional*K*-vector space*V*_{λ}= span{*v*_{1},*v*_{2}}. This becomes a*KQ*-module via the representation Θ_{λ}:*KQ*→End_{K}(*V*_{λ}), where- (a)
Show that for

*λ*≠*μ*the*KQ*-modules*V*_{λ}and*V*_{μ}are not isomorphic. - (b)
Show that for all

*λ*∈*K*the*KQ*-module*V*_{λ}is indecomposable.

- (a)
- 7.11.Let
*K*[*X*] be the polynomial algebra.- (a)
Show that

*K*[*X*] is indecomposable as a*K*[*X*]-module. - (b)
Show that the equivalence in the second version of Fitting’s Lemma (Corollary 7.9) does not hold, by giving a

*K*[*X*]-module endomorphism of*K*[*X*] which is neither invertible nor nilpotent. - (c)
Show that the equivalence in the third version of Fitting’s Lemma (Corollary 7.16) also does not hold for

*K*[*X*].

- (a)
- 7.12.
- (a)
By applying the Artin–Wedderburn theorem, characterize which semisimple

*K*-algebras are local algebras. - (b)
Let

*G*be a finite group such that the group algebra*KG*is semisimple (that is, the characteristic of*K*does not divide |*G*|, by Maschke’s theorem). Deduce that*KG*is not a local algebra, except for the group*G*with one element.

- (a)