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 = (idM − ε)(M), and ε 2 = ε, and (idM − ε)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.
![$$\displaystyle \begin{aligned}x= \varepsilon(m)= \varepsilon^2(m) = \varepsilon (\mathrm{id}_M-\varepsilon)(m) = \varepsilon(m)-\varepsilon^2(m)=0. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equb.png)
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.
![../images/449217_1_En_7_Chapter/449217_1_En_7_Figa_HTML.png](../images/449217_1_En_7_Chapter/449217_1_En_7_Figa_HTML.png)
- (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 = 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 thereforeSimilarly, 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)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 by
We 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 ∈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
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 ⋅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 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).
- (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.
![$$\displaystyle \begin{aligned} \ker(\theta)\subseteq \ker(\theta^2)\subseteq \ker(\theta^3)\subseteq\ldots \subseteq V \end{aligned} $$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equ1.png)
![$$\displaystyle \begin{aligned} V \supseteq \mathrm{im}(\theta)\supseteq \mathrm{im}(\theta^2)\supseteq \mathrm{im}(\theta^3)\supseteq\ldots \end{aligned} $$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equ2.png)
![$$\ker (\theta ^{n_1}) = \ker (\theta ^{n_1+k})$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq19.png)
![$$\mathrm {im}(\theta ^{n_2}) = \mathrm {im}(\theta ^{n_2+k})$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq20.png)
![$$\ker (\theta ^n) \cap \mathrm {im}(\theta ^n) =0$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq21.png)
![$$\ker (\theta ^n) + \mathrm {im}(\theta ^n) = V$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq22.png)
![$$x\in \ker (\theta ^n) \cap \mathrm {im}(\theta ^n)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq23.png)
![$$y\in \ker (\theta ^{2n})=\ker (\theta ^n)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq24.png)
![$$\ker (\theta ^n)\cap \mathrm {im}(\theta ^n)=0$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq25.png)
![$$\displaystyle \begin{aligned} \dim_K V =& \dim_K \ker(\theta^n) + \dim_K \mathrm{im}(\theta^n)\\ =& \dim_K \ker(\theta^n) + \dim_K \mathrm{im}(\theta^n) - \dim_K (\ker(\theta^n)\cap \mathrm{im}(\theta^n))\\ = & \dim_K (\ker(\theta^n) + \mathrm{im}(\theta^n)). \end{aligned} $$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equg.png)
![$$\ker (\theta ^n) + \mathrm {im}(\theta ^n)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq26.png)
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).
- (i)
M is an indecomposable A-module.
- (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.
![$$\displaystyle \begin{aligned}(1_E+a+a^2+\ldots +a^{n-1})(1_E-a)=1_E = (1_E-a)(1_E+a+a^2+\ldots +a^{n-1}),\end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equh.png)
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.
- (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 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.
![$$\displaystyle \begin{aligned}(-a)y = 1_A-ax. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equi.png)
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.
- (a)
Assume A is a local K-algebra. Then the only idempotents in A are 0 and 1A.
- (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 = aε − aε 2 = 0. On the other hand, if ε has a left inverse, say bε = 1A for some b ∈ A, then ε = 1A ε = bε 2 = bε = 1A.
![$$A=\ker (\theta ^n)\oplus \mathrm {im}(\theta ^n)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq28.png)
![$$\displaystyle \begin{aligned}1_A = \varepsilon_1+\varepsilon_2\mbox{ ~with }\varepsilon_1\in \ker(\theta^n),\ \varepsilon_2\in \mathrm{im}(\theta^n). \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equj.png)
![$$A=\ker (\theta ^n)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq29.png)
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 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)
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)
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 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).
![$$\displaystyle \begin{aligned} M_1\oplus \ldots \oplus M_r = M = N_1\oplus \ldots \oplus N_s \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equk.png)
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 EndA(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 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.
![$$\displaystyle \begin{aligned} \mathrm{id}_{N_1} = \nu_1\circ \kappa_1 = \nu_1\circ \mathrm{id}_M\circ \kappa_1 = \sum_{i=1}^r \nu_1\circ e_i\circ \kappa_1. \end{aligned} $$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equ3.png)
![$$\displaystyle \begin{aligned} \phi:= \nu_1\circ e_1\circ \kappa_1 = \nu_1\circ \iota_1\circ \mu_1\circ \kappa_1. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equl.png)
![$$\displaystyle \begin{aligned}M_1 = \mathrm{im}(\mu_1\circ \kappa_1) \oplus \ker(\nu_1\circ \iota_1). \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equm.png)
![$$\ker (\nu _1\circ \iota _1) = 0$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq33.png)
![$$\displaystyle \begin{aligned}\gamma:= \mathrm{id}_M - f_1 + e_1\circ f_1. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equn.png)
![$$\displaystyle \begin{aligned}0 = f_1(0) = (f_1\circ \gamma)(x) = f_1(x)-f_1^2(x)+(f_1\circ e_1\circ f_1)(x) = (f_1\circ e_1\circ f_1)(x). \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equo.png)
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.
![$$\displaystyle \begin{aligned}M = \gamma(M) = \gamma(N_1)\oplus \gamma(N_2) \oplus \ldots \oplus \gamma(N_s) = M_1\oplus N_2\oplus \ldots \oplus N_s. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equp.png)
![$$\displaystyle \begin{aligned}M_2\oplus \ldots \oplus M_r \cong M/M_1 = (M_1\oplus N_2\oplus \ldots \oplus N_s)/M_1 \cong N_2\oplus \ldots \oplus N_s.\end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equq.png)
![$$N_i^{\prime }:= \psi ^{-1}(N_i)$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq34.png)
![$$\displaystyle \begin{aligned}M^{\prime} = N_2^{\prime}\oplus \ldots \oplus N_s^{\prime}. \end{aligned}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_Equr.png)
![$$M_i\cong N^{\prime }_{\sigma (i)} \cong N_{\sigma (i)}$$](../images/449217_1_En_7_Chapter/449217_1_En_7_Chapter_TeX_IEq35.png)
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 minimalsuch 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 − 6X 2 + 12X − 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 Θλ : KQ →EndK(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)