7. Indecomposable Modules

Karin Erdmann1  and Thorsten Holm2
(1)
Mathematical Institute, University of Oxford, Oxford, UK
(2)
Fakult¨at f¨ur Mathematik und Physik, Institut f¨ur Algebra, Zahlentheorie und Diskrete Mathematik, Leibniz Universität Hannover, Hannover, Germany
 

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. (1)

    Every simple module is indecomposable.

     
  2. (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 matrix
    
$$\displaystyle \begin{aligned}\left(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\right) \end{aligned}$$
    with 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. (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. (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.

     
This means that to study indecomposable modules, we should focus on algebras which are not semisimple.

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 = (idM − ε)(M), and ε 2 = ε, and (idMε)2 = idM − 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 EndA(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 EndA(M) does not contain any idempotents except 0 and idM.

Proof.

Assume first that M is indecomposable. Suppose ε ∈EndA(M) is an idempotent, then since idM = ε+(idMε) we have M = ε(M)+(idMε)(M). Moreover, this sum is direct: if x ∈ ε(M) ∩ (idM − ε)(M) then ε(m) = x = (idM − ε)(n) with m, n ∈ M, and then

$$\displaystyle \begin{aligned}x= \varepsilon(m)= \varepsilon^2(m) = \varepsilon (\mathrm{id}_M-\varepsilon)(m) = \varepsilon(m)-\varepsilon^2(m)=0. \end{aligned}$$
Therefore, M = ε(M) ⊕ (idM − ε)(M) is a direct sum of A-submodules. Since M is indecomposable, ε(M) = 0 or (idM − ε)(M) = 0. That is, ε = 0 or else idM − ε = 0, which means ε = idM.

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 EndA(M). By assumption, ε is zero or the identity, which means that U = 0 or V = 0. Hence M is indecomposable.□

Example 7.4.

Submodules, or factor modules of indecomposable modules need not be indecomposable. As an example, consider the path algebra A := KQ of the Kronecker quiver as in Example 1.​13,
../images/449217_1_En_7_Chapter/449217_1_En_7_Figa_HTML.png
  1. (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-submodules
    
$$\displaystyle \begin{aligned}U=\mathrm{span}\{a,b\} = \mathrm{span}\{a\}\oplus \mathrm{span}\{b\}. \end{aligned}$$
    However, 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 = idM 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 therefore
    
$$\displaystyle \begin{aligned}\varepsilon(a) = \varepsilon(ae_1) = a\varepsilon(e_1) = a(\lambda e_1) = \lambda(ae_1) = \lambda a. \end{aligned}$$
    Similarly, since b = be 1 we have ε(b) = λb. We have proved that ε = λ ⋅idM. 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. (2)
    Let N be the A-module with basis 
$$\{v_1,v_1^{\prime },v_2\}$$
where e 1 N has basis 
$$\{v_1,v_1^{\prime }\}$$
and e 2 N has basis {v 2} and where the action of a and b is defined by
    
$$\displaystyle \begin{aligned}av_1=v_2, \ \ av_1^{\prime}=0, \ \ av_2=0 \mbox{ ~and ~} bv_1=0, \ \ bv_1^{\prime}=v_2, \ \ bv_2=0. \end{aligned}$$
    We see that U := span{v 2} is an A-submodule of N. The factor module NU has basis consisting of the cosets of 
$$v_1, v_1^{\prime }$$
, and from the definition, each element in the basis {e 1, e 2, a, b} of A acts by scalar multiplication on NU. As before, this implies that NU 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 ∈EndA(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 
$$\varepsilon (v_1)=\mu v_1+\rho v_1^{\prime }$$
for some μ, ρ ∈ K. Using that bv 1 = 0 and 
$$bv_1^{\prime }=v_2$$
this implies that ρ = 0. Then λv 2 = ε(v 2) = ε(av 1) = (v 1) = μv 2, hence λ = μ and ε(v 1) = λv 1. Similarly, one shows 
$$\varepsilon (v_1^{\prime })=\lambda v_1^{\prime }$$
, that is, we get ε = λ ⋅idN. If ε 2 = ε then ε = 0 or ε = idN.
     

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 dimK M. If dimK M = 1, then M is a simple A-module, hence is an indecomposable A-module, and we are done. So let dimK 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 (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 
$$K=\mathbb {Q}$$
, and let 
$$\mathbb {R}$$
be the set of real numbers. Then 
$$\mathbb {R}$$
is a vector space over 
$$\mathbb {Q}$$
, hence is a 
$$\mathbb {Q}$$
-module. The indecomposable 
$$\mathbb {Q}$$
-modules are the 1-dimensional 
$$\mathbb {Q}$$
-vector spaces. Since 
$$\mathbb {R}$$
has infinite dimension over 
$$\mathbb {Q}$$
, it cannot be a finite direct sum of indecomposable 
$$\mathbb {Q}$$
-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:
  1. (i)

    For all k ≥ 0 we have 
$$\ker (\theta ^n)=\ker (\theta ^{n+k})$$
and im(θ n) = im(θ n+k).

     
  2. (ii)

    
$$~~~~V = \ker (\theta ^n) \oplus \mathrm {im}(\theta ^n)$$
.

     

Proof.

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

(i) We have inclusions of subspaces

$$\displaystyle \begin{aligned} \ker(\theta)\subseteq \ker(\theta^2)\subseteq \ker(\theta^3)\subseteq\ldots \subseteq V \end{aligned} $$
(7.1)
and

$$\displaystyle \begin{aligned} V \supseteq \mathrm{im}(\theta)\supseteq \mathrm{im}(\theta^2)\supseteq \mathrm{im}(\theta^3)\supseteq\ldots \end{aligned} $$
(7.2)
Since V  is finite-dimensional, the ascending chain (7.1) cannot contain infinitely many strict inequalities, so there exists some n 1 ≥ 1 such that 
$$\ker (\theta ^{n_1}) = \ker (\theta ^{n_1+k})$$
for all k ≥ 0. Similarly, for the descending chain (7.2) there is some n 2 ≥ 1 such that 
$$\mathrm {im}(\theta ^{n_2}) = \mathrm {im}(\theta ^{n_2+k})$$
for all k ≥ 0. Setting n as the maximum of n 1 and n 2 proves (i).
(ii) We show 
$$\ker (\theta ^n) \cap \mathrm {im}(\theta ^n) =0$$
and 
$$\ker (\theta ^n) + \mathrm {im}(\theta ^n) = V$$
. Let 
$$x\in \ker (\theta ^n) \cap \mathrm {im}(\theta ^n)$$
, that is, 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 
$$y\in \ker (\theta ^{2n})=\ker (\theta ^n)$$
by part (i), and thus x = θ n(y) = 0. Hence 
$$\ker (\theta ^n)\cap \mathrm {im}(\theta ^n)=0$$
. Now, by the rank-nullity theorem we have

$$\displaystyle \begin{aligned} \dim_K V =&amp; \dim_K \ker(\theta^n) + \dim_K \mathrm{im}(\theta^n)\\ =&amp; \dim_K \ker(\theta^n) + \dim_K \mathrm{im}(\theta^n) - \dim_K (\ker(\theta^n)\cap \mathrm{im}(\theta^n))\\ = &amp; \dim_K (\ker(\theta^n) + \mathrm{im}(\theta^n)). \end{aligned} $$
Hence the sum 
$$\ker (\theta ^n) + \mathrm {im}(\theta ^n)$$
is equal to V  since it is a subspace whose dimension is equal to dimK 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 
$$M=\ker (\theta ^n)\oplus \mathrm {im}(\theta ^n)$$
, 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:
  1. (i)

    M is an indecomposable A-module.

     
  2. (ii)

    Every homomorphism θ ∈EndA(M) is either an isomorphism, or is nilpotent.

     

Proof.

We first assume that statement (i) holds. By Corollary 7.8, for every θ ∈EndA(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.

Corollary 7.9 shows that the endomorphism algebra E := EndA(M) of an indecomposable A-module M has the following property: if a ∈ E is not invertible, then 1E − 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

$$\displaystyle \begin{aligned}(1_E+a+a^2+\ldots +a^{n-1})(1_E-a)=1_E = (1_E-a)(1_E+a+a^2+\ldots +a^{n-1}),\end{aligned}$$
that is, 1E − 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:
  1. (i)

    The set N of elements x  A which do not have a left inverse is a left ideal of A.

     
  2. (ii)

    For all a  A, at least one of a and 1A − 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.

Assume (ii) holds, we show that then 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) = 1A, so that

$$\displaystyle \begin{aligned}(-a)y = 1_A-ax. \end{aligned}$$
We know 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 yN, 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 1A − a has a left inverse in A. If this is false then both a and 1A − a belong to N. By assumption (i), N is closed under addition, therefore 1A ∈ N, which is not true. This contradiction shows that 1A − 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.

  1. (a)

    Assume A is a local K-algebra. Then the only idempotents in A are 0 and 1A.

     
  2. (b)

    Assume A is a finite-dimensional algebra. Then A is local if and only if the only idempotents in A are 0 and 1A.

     

Proof.

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

(b) We must show the converse of (a). Assume that 0 and 1A 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 1A − 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 
$$A=\ker (\theta ^n)\oplus \mathrm {im}(\theta ^n)$$
for some n ≥ 1. So we have a unique expression

$$\displaystyle \begin{aligned}1_A = \varepsilon_1+\varepsilon_2\mbox{ ~with }\varepsilon_1\in \ker(\theta^n),\ \varepsilon_2\in \mathrm{im}(\theta^n). \end{aligned}$$
By Lemma 5.​1, the ε i are idempotents. By assumption, ε 1 = 0 or ε 1 = 1A. Furthermore, by Lemma 5.​1 we have A =  1 ⊕  2. If ε 1 = 0 then A = im(θ n) = Aa n and then a has a left inverse in A (since 1A = ba n for some b ∈ A). Otherwise, ε 1 = 1A and then 
$$A=\ker (\theta ^n)$$
, that is, a n = 0. A computation as in Remark 7.10 then shows that 1A − a has a left inverse, namely 1A + a + a 2 + … + a n−1.□

We will now investigate some examples.

Example 7.15.

  1. (1)

    Let A = K, the one-dimensional K-algebra. Then for every a ∈ A, at least one of a or 1A − 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. (2)

    Let A = M n(K) where n ≥ 2. Let a := E 11, then a and 1A − a do not have a left inverse. Hence A is not local. (We have M 1(K) = K, which is local, by (1).)

     
  3. (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 EndA(M) is a local algebra.

Proof.

Let E := EndA(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 idM. 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 
$$n\in \mathbb {N}$$
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

$$\displaystyle \begin{aligned} M_1\oplus \ldots \oplus M_r = M = N_1\oplus \ldots \oplus N_s \end{aligned}$$

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 iN 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 
$$\mu _i\circ \iota _i=\mathrm {id}_{M_i}$$
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 EndA(M) with 
$${\mathrm {id}}_M = \sum _{i=1}^r e_i$$
.

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 EndA(M) whose sum is the identity idM.

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.

(1) We will find an isomorphism between N 1 and some M i. We have

$$\displaystyle \begin{aligned} \mathrm{id}_{N_1} = \nu_1\circ \kappa_1 = \nu_1\circ \mathrm{id}_M\circ \kappa_1 = \sum_{i=1}^r \nu_1\circ e_i\circ \kappa_1.  \end{aligned} $$
(∗)
The module 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 EndA(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.
Hence at least one of the summands in (∗) is an isomorphism, and we may relabel the M i and assume that ϕ is an isomorphism N 1 → M 1, where

$$\displaystyle \begin{aligned} \phi:= \nu_1\circ e_1\circ \kappa_1 = \nu_1\circ \iota_1\circ \mu_1\circ \kappa_1. \end{aligned}$$
Now we apply Lemma 2.​30 with M = M 1, and N 1 = N = N and j = μ 1 ∘ κ 1 and π = ν 1 ∘ ι 1. We obtain

$$\displaystyle \begin{aligned}M_1 = \mathrm{im}(\mu_1\circ \kappa_1) \oplus \ker(\nu_1\circ \iota_1). \end{aligned}$$
Now, since M 1 is indecomposable and μ 1 ∘ κ 1 is non-zero we have M 1 = im(μ 1 ∘ κ 1) and 
$$\ker (\nu _1\circ \iota _1) = 0$$
. 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.
(2) As a tool for the inductive step, we construct an A-module isomorphism γ : M → M such that γ(N 1) = M 1 and γ(N j) = N j for 2 ≤ j ≤ s. Define

$$\displaystyle \begin{aligned}\gamma:= \mathrm{id}_M - f_1 + e_1\circ f_1. \end{aligned}$$
We first show that 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

$$\displaystyle \begin{aligned}0 = f_1(0) = (f_1\circ \gamma)(x) = f_1(x)-f_1^2(x)+(f_1\circ e_1\circ f_1)(x) = (f_1\circ e_1\circ f_1)(x). \end{aligned}$$
By definition, 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, idM − 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.

(3) We complete the proof of the Krull–Schmidt Theorem. Note that an isomorphism takes a direct sum decomposition to a direct sum decomposition, see Exercise 7.3. By (2) we have

$$\displaystyle \begin{aligned}M = \gamma(M) = \gamma(N_1)\oplus \gamma(N_2) \oplus \ldots \oplus \gamma(N_s) = M_1\oplus N_2\oplus \ldots \oplus N_s. \end{aligned}$$
Using the isomorphism theorem, we have

$$\displaystyle \begin{aligned}M_2\oplus \ldots \oplus M_r \cong M/M_1 = (M_1\oplus N_2\oplus \ldots \oplus N_s)/M_1 \cong N_2\oplus \ldots \oplus N_s.\end{aligned}$$
To apply the induction hypothesis, we need two direct sum decompositions of the same module. Let M  := M 2 ⊕… ⊕ M r. We have obtained an isomorphism ψ : M → N 2 ⊕… ⊕ N s. Let 
$$N_i^{\prime }:= \psi ^{-1}(N_i)$$
, this is a submodule of M , and we have, again by Exercise 7.3, the direct sum decomposition

$$\displaystyle \begin{aligned}M^{\prime} = N_2^{\prime}\oplus \ldots \oplus N_s^{\prime}. \end{aligned}$$
By the induction hypothesis, r − 1 = s − 1 and there is a permutation σ of {1, 2, 3, …, r} with 
$$M_i\cong N^{\prime }_{\sigma (i)} \cong N_{\sigma (i)}$$
for i ≥ 2 (and σ(1) = 1). This completes the proof.□

EXERCISES

  1. 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).

     
  2. 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 submodules
    
$$\displaystyle \begin{aligned}V_i:=\{(\lambda_1,\ldots,\lambda_n)^t\,|\,\lambda_j=0 \mbox{ for all }j&gt;i\} = \mathrm{span}\{e_1,\ldots,e_i\}, \end{aligned}$$
    where 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 iV j (for 0 ≤ j < i ≤ n) are 
$$\frac {n(n+1)}{2}$$
pairwise non-isomorphic T n(K)-modules.
    1. (a)

      Determine the endomorphism algebra 
$$\mathrm {End}_{T_n(K)}(V_{i,j})$$
for all 0 ≤ j < i ≤ n.

       
    2. (b)

      Deduce that each V i,j is an indecomposable T n(K)-module.

       
     
  3. 7.5.

    Recall that for any K-algebra A and every element a ∈ A the map θ a : A → A, bba, 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 
$$d\in \mathbb {N}$$
such that 
$$\mathrm {ker}(\theta _a^{d}) = \mathrm {ker}(\theta _a^{d+j})$$
and 
$$\mathrm {im}(\theta _a^{d}) = \mathrm {im}(\theta _a^{d+j})$$
for all 
$$j\in \mathbb {N}$$
. Moreover, give explicitly the decomposition 
$$T_n(K)=\mathrm {ker}(\theta _a^{d})\oplus \mathrm {im}(\theta _a^{d})$$
(which exists by the Fitting Lemma, see Corollary 7.7):
    1. (i)

      a = E 11;

       
    2. (ii)

      a = E 12 + E 23 + … + E n−1,n;

       
    3. (iii)

      a = E 1n + E 2n + … + E nn.

       
     
  4. 7.6.
    1. (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 
$$m\in \mathbb {N}$$
(up to multiplication by a non-zero scalar in K).

       
    2. (b)
      Which of the following algebras are local?
      1. (i)

        
$$\mathbb {Q}[X]/(X^n-1)$$
, where n ≥ 2;

         
      2. (ii)

        
$$\mathbb {Z}_p[X]/(X^p-1)$$
where p is a prime number;

         
      3. (iii)

        K[X]∕(X 3 − 6X 2 + 12X − 8).

         
       
     
  5. 7.7.
    Which of the following K-algebras A are local?
    1. (i)

      A = T n(K), the algebra of upper triangular n × n-matrices;

       
    2. (ii)

      A = {a = (a ij) ∈ T n(K) | a ii = a jj for all i, j}.

       
     
  6. 7.8.
    For a field K let 
$$K[[X]]:=\{\sum _{i=0}^{\infty } \lambda _iX^i\,|\,\lambda _i\in K\} $$
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.

    1. (a)

      Verify that K[[X]] with these operations becomes a commutative K-algebra.

       
    2. (b)

      Determine the invertible elements in K[[X]].

       
    3. (c)

      Show that K[[X]] is a local algebra.

       
     
  7. 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.

     
  8. 7.10.
    For a field K consider the path algebra KQ of the Kronecker quiver
    ../images/449217_1_En_7_Chapter/449217_1_En_7_Figb_HTML.png
    For 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 Θλ : KQ →EndK(V λ), where
    
$$\displaystyle \begin{aligned}\Theta_{\lambda}(e_1)=\left(\begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 0 \end{array} \right)\,,~ \Theta_{\lambda}(e_2)=\left(\begin{array}{cc} 0 &amp; 0 \\ 0 &amp; 1 \end{array} \right) \end{aligned}$$
    
$$\displaystyle \begin{aligned} \Theta_{\lambda}(a)=\left(\begin{array}{cc} 0 &amp; 0 \\ 1 &amp; 0 \end{array} \right)\,,~ \Theta_{\lambda}(b)=\left(\begin{array}{cc} 0 &amp; 0 \\ \lambda &amp; 0 \end{array} \right).\end{aligned} $$
    1. (a)

      Show that for λ ≠ μ the KQ-modules V λ and V μ are not isomorphic.

       
    2. (b)

      Show that for all λ ∈ K the KQ-module V λ is indecomposable.

       
     
  9. 7.11.
    Let K[X] be the polynomial algebra.
    1. (a)

      Show that K[X] is indecomposable as a K[X]-module.

       
    2. (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.

       
    3. (c)

      Show that the equivalence in the third version of Fitting’s Lemma (Corollary 7.16) also does not hold for K[X].

       
     
  10. 7.12.
    1. (a)

      By applying the Artin–Wedderburn theorem, characterize which semisimple K-algebras are local algebras.

       
    2. (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.