2. Modules and Representations

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

Representation theory studies how algebras can act on vector spaces. The fundamental notion is that of a module, or equivalently (as we shall see), that of a representation. Perhaps the most elementary way to think of modules is to view them as generalizations of vector spaces, where the role of scalars is played by elements in an algebra, or more generally, in a ring.

2.1 Definition and Examples

A vector space over a field K is an abelian group V  together with a scalar multiplication K × V → V , satisfying the usual axioms. If one replaces the field K by a ring R, then one gets the notion of an R-module. Although we mainly deal with algebras over fields in this book, we slightly broaden the perspective in this chapter by defining modules over rings. We always assume that rings contain an identity element.

Definition 2.1.

Let R be a ring with identity element 1R. A left R-module (or just R-module) is an abelian group (M, +) together with a map

$$\displaystyle \begin{aligned}R\times M \to M, \ \ (r, m)\mapsto r\cdot m \end{aligned}$$
such that for all r, s ∈ R and all m, n ∈ M we have
  1. (i)

    (r + s) ⋅ m = r ⋅ m + s ⋅ m;

  2. (ii)

    r ⋅ (m + n) = r ⋅ m + r ⋅ n;

  3. (iii)

    r ⋅ (s ⋅ m) = (rs) ⋅ m;

  4. (iv)

    1 R ⋅ m = m.


Exercise 2.1.

Let R be a ring (with zero element 0R and identity element 1R) and M an R-module with zero element 0M. Show that the following holds for all r ∈ R and m ∈ M:
  1. (i)

    0R ⋅ m = 0M;

  2. (ii)

    r ⋅ 0M = 0M;

  3. (ii)

    − (r ⋅ m) = (−r) ⋅ m = r ⋅ (−m), in particular − m = (−1R) ⋅ m.


Remark 2.2.

Completely analogous to Definition 2.1 one can define right R-modules, using a map M × R → M, (m, r)↦m ⋅ r. When the ring R is not commutative the behaviour of left modules and of right modules can be different; for an illustration see Exercise 2.22. We will consider only left modules, since we are mostly interested in the case when the ring is a K-algebra, and scalars are usually written to the left.

Before dealing with elementary properties of modules we consider a few examples.

Example 2.3.

  1. (1)

    When R = K is a field, then R-modules are exactly the same as K-vector spaces. Thus, modules are a true generalization of the concept of a vector space.

  2. (2)

$$R=\mathbb {Z}$$
, the ring of integers. Then every abelian group can be viewed as a 
$$\mathbb {Z}$$
-module: If n ≥ 1 then n ⋅ a is set to be the sum of n copies of a, and (−n) ⋅ a := −(n ⋅ a), and 
$$0_{\mathbb {Z}} \cdot a = 0$$
. With this, conditions (i) to (iv) in Definition 2.1 hold in any abelian group.

  3. (3)

    Let R be a ring (with 1). Then every left ideal I of R is an R-module, with R-action given by ring multiplication. First, as a left ideal, (I, +) is an abelian group. The properties (i)–(iv) hold even for arbitrary elements in R.

  4. (4)

    A very important special case of (3) is that every ring R is an R-module, with action given by ring multiplication.

  5. (5)
    Suppose M 1, …, M n are R-modules. Then the cartesian product
$$\displaystyle \begin{aligned}M_1\times \ldots \times M_n: = \{ (m_1,\ldots,m_n)\mid m_i \in M_i \} \end{aligned}$$
    is an R-module if one defines the addition and the action of R componentwise, so that
$$\displaystyle \begin{aligned}r\cdot (m_1,\ldots,m_n) := (rm_1,\ldots, rm_n) \mbox{ ~for }r \in R\mbox{ and }m_i \in M_i. \end{aligned}$$
    The module axioms follow immediately from the fact that they hold in M 1, …, M n.

We will almost always study modules when the ring is a K-algebra A. In this case, there is a range of important types of A-modules, which we will now introduce.

Example 2.4.

Let K be a field.
  1. (1)

    If A is a subalgebra of the algebra of n × n-matrices M n(K), or a subalgebra of the algebra EndK(V ) of K-linear maps on a vector space V  (see Example 1.​3), then A has a natural module, which we will now describe.

    1. (i)

      Let A be a subalgebra of M n(K), and let V = K n, the space of column vectors, that is, of n × 1-matrices. By properties of matrix multiplication, multiplying an n × n-matrix by an n × 1-matrix gives an n × 1-matrix, and this satisfies axioms (i) to (iv). Hence V  is an A-module, the natural A-module. Here A could be all of M n(K), or the algebra of upper triangular n × n-matrices, or any other subalgebra of M n(K).

    2. (ii)
      Let V  be a vector space over the field K. Assume that A is a subalgebra of the algebra EndK(V ) of all K-linear maps on V  (see Example 1.​3). Then V  becomes an A-module, where the action of A is just applying the linear maps to the vectors, that is, we set
$$\displaystyle \begin{aligned}A\times V\to V\mbox{ , ~}(\varphi, v) \mapsto \varphi\cdot v:=\varphi(v). \end{aligned}$$
      To check the axioms, let φ, ψ ∈ A and v, w ∈ V , then we have
$$\displaystyle \begin{aligned} (\varphi+\psi)\cdot v = (\varphi+\psi)(v) = \varphi(v) + \psi(v) =\varphi\cdot v + \psi\cdot v \end{aligned}$$
      by the definition of the sum of two maps, and similarly
$$\displaystyle \begin{aligned}\varphi\cdot (v+w) = \varphi(v +w) = \varphi(v) + \varphi(w)= \varphi\cdot v + \varphi\cdot w \end{aligned}$$
      since φ is K-linear. Moreover,
$$\displaystyle \begin{aligned}\varphi\cdot (\psi\cdot v) = \varphi(\psi(v))= (\varphi\psi)\cdot v \end{aligned}$$
      since the multiplication in EndK(V ) is given by composition of maps, and clearly we have 1A ⋅ v = idV(v) = v.
  2. (2)

    Let A = KQ be the path algebra of a quiver Q. For a fixed vertex i, let M be the span of all paths in Q starting at i. Then M = Ae i, which is a left ideal of A and hence is an A-module (see Example 2.3).

  3. (3)
    Let A = KG be the group algebra of a group G. The trivial KG-module has underlying vector space K, and the action of A is defined by
$$\displaystyle \begin{aligned}g\cdot x = x \mbox{ ~for all }g\in G\mbox{ and }x \in K \end{aligned}$$
    and extended linearly to the entire group algebra KG. The module axioms are trivially satisfied.
  4. (4)

    Let B be an algebra and A a subalgebra of B. Then every B-module M can be viewed as an A-module with respect to the given action. The axioms are then satisfied since they even hold for elements in the larger algebra B. We have already used this, when describing the natural module for subalgebras of M n(K), or of EndK(V ).

  5. (5)
    Let A, B be K-algebras and suppose φ : A → B is a K-algebra homomorphism. If M is a B-module, then M also becomes an A-module by setting
$$\displaystyle \begin{aligned}A\times M\to M\mbox{ , ~}(a,m)\mapsto a\cdot m:=\varphi(a)m \end{aligned}$$
    where on the right we use the given B-module structure on M. It is straightforward to check the module axioms.

Exercise 2.2.

Explain briefly why example (4) is a special case of example (5).

We will almost always focus on the case when the ring is an algebra over a field K. However, for some of the general properties it is convenient to see these for rings. In this chapter we will write R and M if we are working with an R-module for a general ring, and we will write A and V  if we are working with an A-module where A is a K-algebra.

Assume A is a K-algebra, then we have the following important observation, namely all A-modules are automatically vector spaces.

Lemma 2.5.

Let K be a field and A a K-algebra. Then every A-module V  is a K-vector space.


Recall from Remark 1.​2 that we view K as a subset of A, by identifying λ ∈ K with λ1A ∈ A. Restricting the action of A on V  gives us a map K × V → V . The module axioms (i)–(iv) from Definition 2.1 are then just the K-vector space axioms for V .□

Remark 2.6.

Let A be a K-algebra. To simplify the construction of A-modules, or to check the axioms, it is usually enough to deal with elements of a fixed K-basis of A, recall Remark 1.​4. Sometimes one can simplify further. For example if A = K[X], it has basis X r for r ≥ 1. Because of axiom (iii) in Definition 2.1 it suffices to define and to check the action of X as this already determines the action of arbitrary basis elements.

Similarly, since an A-module V  is always a K-vector space, it is often convenient (and enough) to define actions on a K-basis of V  and also check axioms using a basis.

2.2 Modules for Polynomial Algebras

In this section we will completely describe the modules for algebras of the form K[X]∕(f) where f ∈ K[X] is a polynomial. We first recall the situation for the case f = 0, that is, modules for the polynomial algebra K[X].

Definition 2.7.

Let K be a field and V  a K-vector space. For any K-linear map α : V → V  we can use this and turn V  into a K[X]-module by setting

$$\displaystyle \begin{aligned}g\cdot v := g(\alpha)(v) = \sum_i \lambda_i \alpha^i(v) \mbox{ ~(for }g=\sum_i \lambda_iX^i\in K[X]\mbox{ and }v\in V\mbox{)}. \end{aligned}$$
Here α i = α ∘… ∘ α is the i-fold composition of maps. We denote this K[X]-module by V α.
Checking the module axioms (i)–(iv) from Definition 2.1 is straightforward. For example, consider condition (iii),

$$\displaystyle \begin{aligned}g\cdot (h\cdot v) = g(\alpha) (h\cdot v) = g(\alpha)(h(\alpha)(v)) = ((gh)(\alpha))(v) = (gh)\cdot v. \end{aligned}$$

Verifying the other axioms is similar and is left as an exercise. Note that, to define a K[X]-module structure on a vector space, one only has to specify the action of X, see Remark 2.6.

Example 2.8.

$$K=\mathbb {R}$$
, and take 
$$V=\mathbb {R}^2$$
, the space of column vectors. Let α : V → V  be the linear map with matrix

$$\displaystyle \begin{aligned}\left(\begin{matrix}0&1\\ 0&0\end{matrix}\right) \end{aligned}$$
with respect to the standard basis of unit vectors of 
$$\mathbb {R}^2$$
. According to Definition 2.7, V  becomes a module for 
$$\mathbb {R}[X]$$
by setting X ⋅ v := α(v). Here α 2 = 0, so if 
$$g =\sum _i \lambda _i X^i\in \mathbb {R}[X]$$
is a general polynomial then

$$\displaystyle \begin{aligned}g\cdot v = g(\alpha)(v) = \sum_i \lambda_i \alpha^i(v) = \lambda_0v + \lambda_1\alpha(v). \end{aligned}$$

The definition of V α is more than just an example. We will now show that every K[X]-module is equal to V α for some K-linear map α.

Proposition 2.9.

Let K be a field and let V  be a K[X]-module. Then V = V α , where α : V  V  is the K-linear map given by α(v) := X  v for v  V.


We first want to show that the map α defined in the statement is K-linear. In fact, for every λ, μ ∈ K and v, w ∈ V  we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha(\lambda v+\mu w) &\displaystyle = &\displaystyle X\cdot (\lambda v+\mu w) = X\cdot (\lambda v) + X\cdot (\mu w) \\ &\displaystyle = &\displaystyle X\cdot (\lambda 1_{K[X]}\cdot v) + X\cdot (\mu 1_{K[X]}\cdot w) \\ &\displaystyle = &\displaystyle (X\lambda1_{K[X]})\cdot v + (X\mu 1_{K[X]})\cdot w \\ &\displaystyle = &\displaystyle (\lambda 1_{K[X]} X)\cdot v + (\mu 1_{K[X]} X)\cdot w \\ &\displaystyle = &\displaystyle \lambda(X\cdot v) + \mu(X\cdot w) = \lambda \alpha(v) + \mu \alpha(w). \end{array} \end{aligned} $$
To prove that V = V α it remains to check that for any polynomial g ∈ K[X] we have g ⋅ v = g(α)(v). In fact, by induction on r one sees that X r ⋅ v = α r(v) for each r, and then one uses linearity to prove the claim.□

The following result relates K[X]-modules with modules for factor algebras K[X]∕I. This is important since for I ≠ 0, these factor algebras are finite-dimensional, and many finite-dimensional algebras occuring ‘in nature’ are of this form. Therefore, this result will be applied many times throughout the book. Recall from basic algebra that K[X] is a principal ideal domain, so that every ideal of K[X] can be generated by one element. Consider now an ideal of K[X] of the form (f) = K[X]f for some polynomial f ∈ K[X]. We assume that f has positive degree, to ensure that K[X]∕(f) is an algebra (that is, contains a non-zero identity element).

Theorem 2.10.

Let K be a field and f  K[X] a non-constant polynomial, and let A be the factor algebra A = K[X]∕(f). Then there is a bijection between the set of A-modules and the set of those K[X]-modules V α which satisfy f(α) = 0.


We set I = (f). Suppose first that V  is an A-module. Then we can view V  as a module for K[X] by setting

$$\displaystyle \begin{aligned} g\cdot v = (g+I)\cdot v \mbox{ ~for all }v\in V\mbox{ and }g\in K[X]. \end{aligned}$$
(Note that this is a special case of Example 2.4, for the algebra homomorphism K[X] → A, gg + I.) Then as a K[X]-module, V = V α, where α is the linear map vX ⋅ v, by Proposition 2.9. It follows that for every v ∈ V  we have

$$\displaystyle \begin{aligned}f(\alpha)(v)= f\cdot v =(f +I)\cdot v =0\cdot v =0 \end{aligned}$$
since f ∈ I, that is, f(α) = 0, as desired.
Conversely, consider a K[X]-module V α where f(α) = 0. Then we view V α as a module for A, by setting

$$\displaystyle \begin{aligned} (g+I)\cdot v:= g(\alpha)(v) \mbox{ ~for all }g\in K[X]\mbox{ and }v\in V. \end{aligned}$$
This is well-defined, that is, independent of the representatives of the cosets: if g + I = h + I then g − h ∈ I = (f) and thus g − h = pf for some polynomial p ∈ K[X]. Therefore (g − h)(α) = (pf)(α) = p(α)f(α) = 0 by assumption, and hence g(α) = h(α). Finally, one checks that the above action satisfies the module axioms from Definition 2.1.

Since we only changed the point of view, and did not do anything to the vector space and the action as such, the two constructions above are inverse to each other and hence give bijections as claimed in the theorem.□

Example 2.11.

Let K be a field.
  1. (1)

    Consider the K-algebras K[X]∕(X n), where 
$$n\in \mathbb {N}$$
. By Theorem 2.10, the K[X]∕(X n)-modules are in bijection with those K[X]-modules V α such that α n = 0. In particular, the 1-dimensional K[X]∕(X n)-modules correspond to scalars α ∈ K such that α n = 0. But for every field K this has only one solution α = 0, that is, there is precisely one 1-dimensional K[X]∕(X n)-module V 0 = span{v} with zero action X ⋅ v = 0. Note that this is independent of the field K.

  2. (2)

    Let A = K[X]∕(X n − 1), where n ≥ 1. By Theorem 2.10, we have that A-modules are in bijection with the K[X]-modules V α such that α n = idV. In particular, a 1-dimensional A-module is given by the K[X]-modules V α, where V = span{v}≅K is a 1-dimensional vector space and α : K → K satisfies α n = idK. That is, α ∈ K is an n-th root of unity. Hence the number of 1-dimensional A-modules depends on how many n-th roots of unity K contains.

$$K=\mathbb {C}$$
, we have n possibilities, namely α = (e 2πin)j for j = 0, 1, …, n − 1.

$$K=\mathbb {R}$$
, then we only get one or two 1-dimensonal A-modules. If n is even then α = 1 or − 1, and if n is odd we only have α = 1.

  3. (3)

    In general, the K-algebra K[X]∕(f) has a 1-dimensional module if and only if the polynomial f has a root in K. So, for example, 
$$\mathbb {R}[X]/(X^2+1)$$
does not have a 1-dimensional module.

  4. (4)

    Let C n be a cyclic group of order n. We have seen in Example 1.​27 that the group algebra KC n is isomorphic to the factor algebra K[X]∕(X n − 1). So we get information on KC n from example (2).


2.3 Submodules and Factor Modules

In analogy to the vector space constructions, we expect there to be ‘submodules’ and ‘factor modules’ of R-modules. A submodule of an R-module M will be a subset U of M which itself is a module with the operations inherited from M. If so, then one can turn the factor group MU into an R-module.

Definition 2.12.

Let R be a ring and let M be some R-module. An R-submodule U of M is a subgroup (U, +) which is closed under the action of R, that is, r ⋅ u ∈ U for all r ∈ R and u ∈ U.

Example 2.13.

  1. (1)

    Let R = K be a field. Then K-modules are just K-vector spaces, and K-submodules are just K-subspaces.

  2. (2)

$$R=\mathbb {Z}$$
be the ring of integers. Then 
$$\mathbb {Z}$$
-modules are nothing but abelian groups, and 
$$\mathbb {Z}$$
-submodules are just (abelian) subgroups.

  3. (3)

    Let R be a ring, considered as an R-module, with action given by ring multiplication. Then the R-submodules of R are precisely the left ideals of R.

  4. (4)

    Every R-module M has the trivial submodules {0} and M. We often just write 0 for the trivial submodule consisting only of the zero element.

  5. (5)
    Let R be a ring and M an R-module. For every m ∈ M the subset
$$\displaystyle \begin{aligned}Rm:=\{r\cdot m\,|\,r\in R\} \end{aligned}$$
    is an R-submodule of M, the submodule generated by m.
  6. (6)

    Assume M is an R-module, and U, V  are R-submodules of M. Then the intersection U ∩ V  is a submodule of M. Furthermore, let U + V := {u + vu ∈ U, v ∈ V }. Then U + V  is a submodule of M.


Exercise 2.3.

Prove the statements in Example 2.13 (6).

For modules of K-algebras, we have some specific examples of submodules.

Example 2.14.

Let K be a field.
  1. (1)
    Assume A = K[X]. We consider submodules of an A-module V α. Recall V  is a vector space, and α is a linear transformation of V , and the action of A is given by
$$\displaystyle \begin{aligned}g\cdot v = g(\alpha)(v) \ \ (v\in V, g\in K[X]). \end{aligned}$$
    Then an A-submodule of V α is the same as a subspace U of V  such that α(U) ⊆ U. For example, the kernel U = ker(α), and the image U = im(α), are A-submodules of V α.
  2. (2)

    Consider the matrix algebra M n(K), and let K n be the natural A-module, see Example 2.4. We claim that the only M n(K)-submodules of K n are the trivial submodules 0 and K n. In fact, let U ⊆ K n be a non-zero M n(K)-submodule, then there is some u = (u 1, …, u n)t ∈ U ∖{0}, say u j ≠ 0 for some j. For every i ∈{1, …, n} consider the matrix unit E ij ∈ M n(K) with entry 1 at position (i, j) and 0 otherwise. Then (u j)−1 E ij ⋅ u is the ith unit vector and it is in U since U is an M n(K)-submodule. So U contains all unit vectors and hence U = K n, as claimed.

  3. (3)

    Let A = T n(K), the algebra of upper triangular matrices. We can also consider K n as an A-module. Then K n has non-trivial submodules, for example there is a 1-dimensional submodule, spanned by (1, 0, …, 0)t. Exercise 2.14 determines all T n(K)-submodules of K n.

  4. (4)

    Let Q be a quiver and A = KQ, the path algebra of Q. For any r ≥ 1, let A r be the subspace of A spanned by paths of length ≥ r. Then A r is a submodule of the A-module A. We have seen Ae i if i is a vertex of Q, this is also a submodule of A. Then we also have the submodule (Ae i)r := Ae i ∩ A r, by Example 2.13.

  5. (5)

    Consider the 2-dimensional 
$$\mathbb {R}$$
$$A_0=\mathrm {span}\{1_{A_0},\tilde {b}\}$$
$$\tilde {b}^2=0$$
as in Sect. 1.​4, as an A-module. The 1-dimensional subspace spanned by 
$$\tilde {b}$$
is an A 0-submodule of A 0. Alternatively, 
$$A_0\cong \mathbb {R}[X]/(X^2)$$
and the subspace span{X + (X 2)} is an 
$$\mathbb {R}[X]/(X^2)$$
-submodule of 
$$\mathbb {R}[X]/(X^2)$$

    On the other hand, consider the algebra 
$$A_1 =\mathrm {span}\{1_{A_1},\tilde {b}\}$$
$$\tilde {b}^2=1_{A_1}$$
in the same section, then the subspace spanned by 
$$\tilde {b}$$
is not a submodule. But the space spanned by 
$$\tilde {b} -1_{A_1}$$
is a submodule. Alternatively, 
$$A_1\cong \mathbb {R}[X]/(X^2-1)$$
; here the subspace U 1 := span{X + (X 2 − 1)} is not a submodule since
$$\displaystyle \begin{aligned}X\cdot (X+(X^2-1)) = 1 +(X^2-1)\not\in U_1, \end{aligned}$$
    but the subspace U 2 := span{X − 1 + (X 2 − 1)} is a submodule since
$$\displaystyle \begin{aligned}X\cdot (X-1+(X^2-1)) = 1-X +(X^2-1)\in U_2. \end{aligned}$$

Exercise 2.4.

$$A = \mathbb {R}[X]/(X^2+1)$$
(which is the algebra A −1 in Sect. 1.​4). Why does A as an A-module not have any submodules except 0 and A?

Sometimes a module can be broken up into ‘smaller pieces’; the fundamental notion for such phenomena is that of the direct sum of submodules. Very often we will only need finite direct sums, but when dealing with semisimple modules we will also need arbitrary direct sums. For clarity, we will give the definition in both situations separately.

Definition 2.15.

Let R be a ring and let M be an R-module.
  1. (a)
    Let U 1, U 2, …, U t be R-submodules of M. We say that M is the direct sum of U 1, …, U t, denoted M = U 1 ⊕ U 2 ⊕… ⊕ U t, if the following two conditions are satisfied:
    1. (i)

      M = U 1 + U 2 + … + U t, that is, every element of M can be expressed as a sum of elements from the submodules U i.

    2. (ii)

      For every j with 1 ≤ j ≤ t we have U j ∩∑ij U i = 0.

  2. (b)
    We say that M is the direct sum of a family (U i)iI of R-submodules (where I is an arbitrary index set), denoted M =⊕iI U i, if the following two conditions are satisfied:
    1. (i)

      M =∑iI U i, that is, every element of M can be expressed as a finite sum of elements from the submodules U i.

    2. (ii)

      For every j ∈ I we have U j ∩∑ij U i = 0.


Remark 2.16.

  1. (1)

    When the ring is a K-algebra A, conditions (i) and (ii) in the above definition state that M is, as a vector space, the direct sum of the subspaces U 1, …, U t (or U i where i ∈ I). One should keep in mind that each U i must be an A-submodule of M.

  2. (2)

    Note that for R = K a field, every K-module (that is, K-vector space) can be decomposed as a direct sum of 1-dimensional K-submodules (that is, K-subspaces); this is just another formulation of the fact that every K-vector space has a basis.


Given an algebra A, expressing A as an A-module as a direct sum of submodules has numerous applications, we will see this later. The following exercises are examples.

Exercise 2.5.

Let A = M n(K), the algebra of n × n-matrices over a field K, considered as an A-module. For any i ∈{1, …, n} we define C i ⊆ A to be the set of matrices which are zero outside the i-th column. Show that C i is an A-submodule of A, and that

$$\displaystyle \begin{aligned}A= C_1\oplus C_2\oplus\ldots \oplus C_n \end{aligned}$$
is a direct sum of these submodules.

Exercise 2.6.

Let A = KQ, the path algebra of some quiver. Suppose the vertices of Q are 1, 2, …, n. Recall that the trivial paths e i ∈ KQ are orthogonal idempotents (that is, 
and e i e j = 0 for i ≠ j), and that e 1 + e 2 + … + e n = 1A. Show that as an A-module, A = Ae 1 ⊕ Ae 2 ⊕… ⊕ Ae n, the direct sum of A-submodules.

Recall that we have seen direct products of finitely many modules for a ring (in Example 2.3). These products are necessary to construct a new module from given modules, which may not be related in any way. We will later need arbitrary direct products, and also some important submodules therein.

Definition 2.17.

Let R be a ring and let (M i)iI be a family of R-modules, where I is some index set.
  1. (a)
    The cartesian product
$$\displaystyle \begin{aligned}\prod_{i\in I} M_i = \{ (m_i)_{i\in I}\,|\,m_i\in M_i\mbox{ for all }i\in I\} \end{aligned}$$
    becomes an R-module with componentwise addition and R-action. This R-module is called the direct product of the family (M i)iI of R-modules.
  2. (b)
    The subset
$$\displaystyle \begin{aligned}\bigoplus_{i\in I} M_i = \{ (m_i)_{i\in I}\,|\,m_i\in M_i\mbox{ for all }i\in I\mbox{, only finitely many }m_i\mbox{ non-zero}\} \end{aligned}$$
    is an R-submodule of the direct product ∏iI M i, called the direct sum of the family (M i)iI of R-modules.

Note that we now have two notions of a direct sum. In Definition 2.15 it is assumed that in M =⊕iI U i the U i are submodules of the given R-module M, whereas in Definition 2.17 the modules M i in ⊕iI M i are not related and a priori are not contained in some given module. (Some books distinguish between these two constructions, calling them an ‘internal’ and ‘external’ direct sum.)

We now come to the important construction of factor modules. Suppose U is a submodule of an R-module M, then one knows from basic algebra that the cosets

$$\displaystyle \begin{aligned}M/U:= \{ m+U\mid m\in M\} \end{aligned}$$
form an abelian group, with addition (m + U) + (n + U) = m + n + U. In the case, when M is an R-module and U is an R-submodule, this is actually an R-module in a natural way.

Proposition 2.18.

Let R be a ring, let M be an R-module and U an R-submodule of M. Then the cosets MU form an R-module if one defines

$$\displaystyle \begin{aligned}r\cdot (m+U) := rm + U \mathit{\mbox{ ~for all }}r\in R\mathit{\mbox{ and }}m\in M. \end{aligned}$$

This module is called the factor module of M modulo U.


One has to check that the R-action is well-defined, that is, independent of the choice of representatives. If m + U = m  + U then m − m ∈ U and then also rm − rm  = r(m − m ) ∈ U since U is an R-submodule. Therefore rm + U = rm  + U. Finally, the module axioms are inherited from those for M.□

Example 2.19.

  1. (1)

    Let R be a ring and I some left ideal of R, then RI is a factor module of the R-module R. For example, if 
$$R=\mathbb {Z}$$
then every ideal is of the form 
$$I = \mathbb {Z} d$$
$$d\in \mathbb {Z}$$
, and this gives the well-known 
$$\mathbb {Z}$$
$$\mathbb {Z}/\mathbb {Z} d$$
of integers modulo d.

  2. (2)

    Let A = KQ, the path algebra of a quiver Q, and let M = Ae i where i is a vertex of Q. This has the submodule (Ae i)≥1 = Ae i ∩ A ≥1 (see Example 2.14). The factor module Ae i∕(Ae i)≥1 is 1-dimensional, spanned by the coset of e i. Note that Ae i may be infinite-dimensional.


2.4 Module Homomorphisms

We have introduced modules as a generalization of vector spaces. In this section we introduce the suitable maps between modules; these ‘module homomorphisms’ provide a direct generalization of the concept of a K-linear map of vector spaces.

Definition 2.20.

Suppose R is a ring, and M and N are R-modules. A map ϕ : M → N is an R-module homomorphism if for all m, m 1, m 2 ∈ M and r ∈ R we have
  1. (i)

    ϕ(m 1 + m 2) = ϕ(m 1) + ϕ(m 2);

  2. (ii)

    ϕ(rm) = (m).

An isomorphism of R-modules is an R-module homomorphism which is also bijective. The R-modules M and N are then called isomorphic; notation: MN. Note that if ϕ is an isomorphism of modules then so is the inverse map; the proof is analogous to Exercise 1.​14.

Remark 2.21.

Assume that the ring in Definition 2.20 is a K-algebra A (for some field K).
  1. (1)
    The above definition also says that every A-module homomorphism ϕ must be K-linear. Indeed, recall that we identified scalars λ ∈ K with elements λ1A in A. Then we have for λ, μ ∈ K and m 1, m 2 ∈ M that
$$\displaystyle \begin{aligned}\phi(\lambda m_1 + \mu m_2) &= \phi((\lambda 1_A)m_1 + (\mu 1_A)m_2) \\ &= (\lambda 1_A)\phi(m_1) + (\mu 1_A)\phi(m_2) \\ &= \lambda \phi(m_1) + \mu \phi(m_2). \end{aligned} $$
  2. (2)

    It suffices to check condition (ii) in Definition 2.20 for elements r ∈ A in a fixed basis, or just for elements which generate A, such as r = X in the case when A = K[X].


Exercise 2.7.

Suppose V  is an A-module where A is a K-algebra. Let EndA(V ) be the set of all A-module homomorphisms V → V , which is by Remark 2.21 a subset of EndK(V ). Check that it is a K-subalgebra.

Example 2.22.

Let R be a ring.
  1. (1)

    Suppose U is a submodule of an R-module M, then the ‘canonical map’ π : M → MU defined by π(m) = m + U is an R-module homomorphism.

  2. (2)
    Suppose M is an R-module, and m ∈ M. Then there is an R-module homomorphism
$$\displaystyle \begin{aligned}\phi: R\to M, \ \ \phi(r) = rm. \end{aligned}$$
    This is very common, and we call it a ‘multiplication homomorphism’. There is a general version of this, also very common. Namely, suppose m 1, m 2, …, m n are given elements in M. Now take the R-module R n := R × R ×… × R, and define
$$\displaystyle \begin{aligned}\psi: R^n \to M, \ \ \psi(r_1, r_2, \ldots, r_n) = r_1m_1 + r_2m_2 + \ldots + r_nm_n. \end{aligned}$$
    You are encouraged to check that this is indeed an R-module homomorphism.
  3. (3)
    Suppose M = M 1 ×… × M r, the direct product of R-modules M 1, …, M r. Then for every i ∈{1, …, r} the projection map
$$\displaystyle \begin{aligned}\pi_i:M\to M_i\mbox{ , ~}(m_1,\ldots,m_r)\mapsto m_i \end{aligned}$$
    is an R-module homomorphism. Similarly, the inclusion maps into the i-th coordinate
$$\displaystyle \begin{aligned}\iota_i: M_i \to M\mbox{ , ~}m_i \mapsto (0,\ldots,0,m_i,0,\ldots,0) \end{aligned}$$
    are R-module homomorphisms. We recommend to check this as well.

    Similarly, if M = U ⊕ V , the direct sum of two submodules U and V , then the projection maps, and the inclusion maps, are R-module homomorphisms.


Now consider the case when the ring is an algebra, then we have several other types of module homomorphisms.

Example 2.23.

Let K be a field.
  1. (1)
    Consider the polynomial algebra A = K[X]. For a K-linear map between A-modules to be an A-module homomorphism it suffices that it commutes with the action of X. We have described the A-modules (see Sect. 2.2), so let V α and W β be A-modules. An A-module homomorphism is a K-linear map θ : V → W such that θ(Xv) = (v) for all v ∈ V . On V α, the element X acts by α, and on W β, the action of X is given by β. So this means
$$\displaystyle \begin{aligned}\theta(\alpha(v)) = \beta(\theta(v)) \mbox{ ~for all }v \in V. \end{aligned}$$
    This holds for all v, so we have θ ∘ α = β ∘ θ.

    Conversely, if θ : V → W is a linear map such that θ ∘ α = β ∘ θ, then θ defines a module homomorphism. In particular, V αW β are isomorphic A-modules if and only if there is an invertible linear map θ such that θ −1 ∘ β ∘ θ = α. This condition means that α and β are similar as linear maps.

  2. (2)

    Let A = K[X] and let V α, W β be 1-dimensional A-modules. In this case, α and β are scalar multiples of the identity map. Then by the previous result, V α and W β are isomorphic if and only if α = β. This follows from the fact that, since β is a scalar multiple of the identity map, it commutes with θ.

  3. (3)

    Let G = C n be a cyclic group of order n. Then we have seen in Example 2.11 that the group algebra 
$$\mathbb {C}G\cong \mathbb {C}[X]/(X^n-1)$$
has n one-dimensional modules given by multiplication with the scalars (e 2πin)j for j = 0, 1, …, n − 1. These are pairwise non-isomorphic, by part (2) above. Thus, 
$$\mathbb {C}G$$
has precisely n one-dimensional modules, up to isomorphism.


Exercise 2.8.

Let A = KQ, where Q is a quiver. We have seen that A has for each vertex i of Q a 1-dimensional module S i := Ae i∕(Ae i)≥1 (see Example 2.19). Show that if i ≠ j then S i and S j are not isomorphic.

In analogy to the isomorphism theorem for linear maps, or group homomorphisms, there are also isomorphism theorems for module homomorphisms.

Theorem 2.24 (Isomorphism Theorems).

Let R be a ring. Then the following hold.
  1. (a)
    Suppose ϕ : M  N is an R-module homomorphism. Then the kernel ker(ϕ) is an R-submodule of M and the image im(ϕ) is an R-submodule of N. Moreover, we have an isomorphism of R-modules
$$\displaystyle \begin{aligned}M/{\mathrm{ker}}(\phi)\cong {\mathrm{im}}(\phi). \end{aligned}$$
  2. (b)
    Suppose U, V  are submodules of an R-module M, then the sum U + V  and the intersection U  V  are also R-submodules of M. Moreover,
$$\displaystyle \begin{aligned}U/(U\cap V)\cong (U+V)/V \end{aligned}$$
    are isomorphic as R-modules.
  3. (c)
    Let M be an R-module. Suppose U  V  M are R-submodules, then VU is an R-submodule of MU, and
$$\displaystyle \begin{aligned}(M/U)/(V/U) \cong M/V \end{aligned}$$
    are isomorphic as R-modules.


Readers who have seen the corresponding theorem for abelian groups may apply this, and just check that all maps involved are R-module homomorphisms. But for completeness, we give a slightly more detailed proof here.

(a) Since ϕ is in particular a homomorphism of (abelian) groups, we know (or can easily check) that ker(ϕ) is a subgroup of M. Moreover, for every m ∈ker(ϕ) and r ∈ R we have

$$\displaystyle \begin{aligned} \phi(rm) = r\phi(m) = r\cdot 0 = 0 \end{aligned}$$
and rm ∈ker(ϕ). Similarly one checks that the image im(ϕ) is a submodule of N. For the second statement we consider the map

$$\displaystyle \begin{aligned}\psi: M/\mathrm{ker}(\phi) \to \mathrm{im}(\phi), \ \ m+\mathrm{ker}(\phi)\mapsto \phi(m). \end{aligned}$$
This map is well-defined and injective: in fact, for any m 1, m 2 ∈ M we have

$$\displaystyle \begin{aligned} m_1+\mathrm{ker}(\phi)=m_2+\mathrm{ker}(\phi) \Longleftrightarrow m_1-m_2\in \mathrm{ker}(\phi) \Longleftrightarrow \phi(m_1) = \phi(m_2). \end{aligned}$$
By definition, ψ is surjective. It remains to check that this map is an R-module homomorphism. For every m 1, m 2, m ∈ M and r ∈ R we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi((m_1+\mathrm{ker}(\phi))+(m_2+\mathrm{ker}(\phi))) &\displaystyle = &\displaystyle \phi(m_1+m_2) = \phi(m_1) + \phi(m_2) \\ &\displaystyle = &\displaystyle \psi(m_1+\mathrm{ker}(\phi))+\psi(m_2+\mathrm{ker}(\phi)), \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi(r(m+\mathrm{ker}(\phi))) = \phi(rm) = r\phi(m) = r\psi(m+\mathrm{ker}(\phi)). \end{array} \end{aligned} $$
(b) We have already seen in Example 2.13 that U + V  and U ∩ V  are submodules. Then we consider the map

$$\displaystyle \begin{aligned}\psi: U \to (U+V)/V\mbox{ , ~} u\mapsto u+V. \end{aligned}$$
From the addition and R-action on factor modules being defined on representatives, it is easy to check that ψ is an R-module homomorphism. Since every coset in (U + V )∕V  is of the form u + v + V = u + V , the map ψ is surjective. Moreover, it follows directly from the definition that ker(ψ) = U ∩ V . So part (a) implies that

$$\displaystyle \begin{aligned}U/(U\cap V) \cong (U+V)/V. \end{aligned}$$
(c) That VU is an R-submodule of MU follows directly from the fact that V  is an R-submodule of M. We then consider the map

$$\displaystyle \begin{aligned}\psi: M/U\to M/V\mbox{ , ~}m+U\mapsto m+V. \end{aligned}$$
Note that this map is well-defined since U ⊆ V  by assumption. One checks that ψ is an R-module homomorphism. By definition, ψ is surjective, and the kernel consists precisely of the cosets of the form m + U with m ∈ V , that is, ker(ψ) = VU. So part (a) implies that

$$\displaystyle \begin{aligned} (M/U)/(V/U) \cong M/V, \end{aligned}$$
as claimed.□

Example 2.25.

Let R be a ring and M an R-module. For any m ∈ M consider the R-module homomorphism

$$\displaystyle \begin{aligned}\phi:R\to M\mbox{ , ~}\phi(r)=rm \end{aligned}$$
from Example 2.22. The kernel

$$\displaystyle \begin{aligned} \mathrm{Ann}_R(m):=\{r\in R\,|\,rm=0\} \end{aligned}$$
is called the annihilator of m in R. By the isomorphism theorem we have that

$$\displaystyle \begin{aligned}R/\mathrm{Ann}_R(m) \cong \mathrm{im}(\phi) = Rm, \end{aligned}$$
that is, the factor module is actually isomorphic to the submodule of M generated by m; this has appeared already in Example 2.13.

In the isomorphism theorem we have seen that factor modules occur very naturally in the context of module homomorphisms. We now describe the submodules of a factor module. This so-called submodule correspondence is very useful, as it allows one to translate between factor modules and modules. This is based on the following observation:

Proposition 2.26.

Let R be a ring and ϕ : M  N an R-module homomorphism. Then for every R-submodule WN the preimage ϕ −1(W) := {m  M | ϕ(m)  ∈ W} is an R-submodule of M, which contains the kernel of ϕ.


We first show that ϕ −1(W) is a subgroup. It contains the zero element since ϕ(0) = 0 ∈ W. Moreover, if m 1, m 2 ∈ ϕ −1(W), then

$$\displaystyle \begin{aligned}\phi(m_1\pm m_2) = \phi(m_1)\pm \phi(m_2) \in W \end{aligned}$$
since W is a subgroup of N, that is, m 1 ± m 2 ∈ ϕ −1(W). Finally, let r ∈ R and m ∈ ϕ −1(W). Then ϕ(rm) = (m) ∈ W since W is an R-submodule, that is, rm ∈ ϕ −1(W). The kernel of ϕ is mapped to zero, and 0 ∈ W, hence the kernel of ϕ is contained in ϕ −1(W).□

Example 2.27.

Let R be a ring and M an R-module, and let U be a submodule of M with factor module MU. Then we have the natural homomorphism π : M → MU, π(m) = m + U. Proposition 2.26 with ϕ = π shows that for every submodule W of MU, the preimage under π is a submodule of M containing U. Explicitly, this is the module

$$\displaystyle \begin{aligned} \widehat{W}:= \pi^{-1}(W) = \{ m\in M\mid m+U \in W\}. \end{aligned}$$

This construction leads to the following submodule correspondence:

Theorem 2.28.

Let R be a ring. Suppose M is an R-module and U is an R-submodule of M. Then the map 
$$W\mapsto \widehat {W}$$
induces a bijection, inclusion preserving, between the set 
$$\mathcal {F}$$
of R-submodules of MU and the set 
$$\mathcal {S}$$
of R-submodules of M that contain U. Its inverse takes V  in 
$$\mathcal {S}$$
$$V/U \in \mathcal {F}$$


If W is a submodule of MU then 
$$\widehat {W}$$
belongs to the set 
$$\mathcal {S}$$
, by Example 2.27. Take a module V  which belongs to 
$$\mathcal {S}$$
, then VU is in 
$$\mathcal {F}$$
, by part (c) of Theorem 2.24. To show that this gives a bijection, we must check that
  1. (i)

    if W is in 
$$\mathcal {F}$$
$$\widehat {W}/U = W$$

  2. (ii)

    if V  is in 
$$\mathcal {S}$$
$$\widehat {(V/U)} = V$$


To prove (i), let 
$$m\in \widehat {W}$$
, then by definition m + U ∈ W so that 
$$\widehat {W}/U \subseteq W$$
. Conversely if w ∈ W then w = m + U for some m ∈ M. Then 
$$m\in \widehat {W}$$
and therefore 
$$w\in \widehat {W}/U$$

To prove (ii), note that 
$$\widehat {(V/U)} = \{ m\in M\mid m+U \in V/U\}$$
. First, let 
$$m\in \widehat {(V/U)}$$
, that is, m + U ∈ VU. Then m + U = v + U for some v ∈ V , and therefore m − v ∈ U. But U ⊆ V  and therefore m − v ∈ V , and it follows that m = (m − v) + v ∈ V. This shows that 
$$\widehat {(V/U)} \subseteq V$$
. Secondly, if v ∈ V  then v + U ∈ VU, so that 
$$v\in \widehat {(V/U)}$$
. This completes the proof of (ii). It is clear from the constructions that they preserve inclusions.□

Example 2.29.

For a field K, we consider the K-algebra A = K[X]∕(X n) as an A-module. We apply Theorem 2.28 in the case R = M = K[X] and U = (X n); then the K[X]-submodules of A are in bijection with the K[X]-submodules of K[X] containing the ideal (X n). The K[X]-submodules of K[X] are precisely the (left) ideals. Since K[X] is a principal ideal domain every ideal is of the form (g) = K[X]g for some polynomial g ∈ K[X]; moreover, (g) contains the ideal (X n) if and only if g divides X n. Hence there are precisely n + 1 different K[X]-submodules of A = K[X]∕(X n), namely all

$$\displaystyle \begin{aligned}(X^j)/(X^n) = \{ X^jh + (X^n)\mid h\in K[X]\} \ \ (0\leq j\leq n). \end{aligned}$$
Note that the K[X]-action on A is the same as the A-action on A (both given by multiplication in K[X]); thus the K[X]-submodules of A are the same as the A-submodules of A and the above list gives precisely the submodules of A as an A-module.
Alternatively, we know that as a K[X]-module, A = V α, where V = A as a vector space, and α is the linear map which is given as multiplication by X (see Sect. 2.2). The matrix of α with respect to the basis {1 + (X n), X + (X n), …, X n−2 + (X n), X n−1 + (X n)} is the Jordan block J n(0) of size n with zero diagonal entries, that is,

$$\displaystyle \begin{aligned}J_n(0)= \left(\begin{matrix} 0 & 0 & \ldots & \ldots & 0\\ 1 & 0 & & & \vdots \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & & \ddots & 0 &0\\ 0 & \ldots & 0 & 1 &0\end{matrix}\right). \end{aligned}$$
A submodule is a subspace which is invariant under this matrix. With this, we also get the above description of submodules.

Exercise 2.9.

Check the details in the above example.

Often, properties of module homomorphisms already give rise to direct sum decompositions. We give an illustration of this, which will actually be applied twice later.

Lemma 2.30.

Let A be a K-algebra, and let M, N, N be non-zero A-modules. Suppose there are A-module homomorphisms j : N  M and π : M  N such that the composition π  j : N  N is an isomorphism. Then j is injective and π is surjective, and M is the direct sum of two submodules,

$$\displaystyle \begin{aligned} M= \mathrm{im}(j) \oplus \ker(\pi). \end{aligned}$$


The first part is clear. We must show that 
$$\mathrm {im}(j) \cap \ker (\pi ) =0$$
and that 
$$M=\mathrm {im}(j) + \ker (\pi )$$
, see Definition 2.15.

$$w\in \mathrm {im}(j) \cap \ker (\pi )$$
, so that w = j(n) for some n ∈ N and π(w) = 0. Then 0 = π(w) = (π ∘ j)(n) from which it follows that n = 0 since π ∘ j is injective. Clearly, then also w = 0, as desired. This proves that the intersection 
$$\mathrm {im}(j)\cap \ker (\pi )$$
is zero.

Let ϕ : N → N be the inverse of π ∘ j, so that we have 
$$\pi \circ j\circ \phi = \mathrm {id}_{N^{\prime }}$$
. Take w ∈ M then

$$\displaystyle \begin{aligned}w = (j\circ \phi\circ \pi)(w) + (w- (j\circ \phi\circ \pi)(w)). \end{aligned}$$
The first summand belongs to im(j), and the second summand is in 
$$\ker (\pi )$$

$$\displaystyle \begin{aligned}\pi(w-(j\circ \phi\circ \pi)(w)) = \pi(w) - (\pi\circ j\circ \phi\circ \pi)(w) = \pi(w) - \pi(w) =0.\end{aligned}$$
This proves that 
$$w \in \mathrm {im}(j) + \ker (\pi )$$
, and hence 
$$M = \mathrm {im}(j) + \ker (\pi )$$

2.5 Representations of Algebras

In basic group theory one learns that a group action of a group G on a set Ω is ‘the same’ as a group homomorphism from G into the group of all permutations of the set Ω.

In analogy, let A be a K-algebra; as we will see, an A-module V  ‘is the same’ as an algebra homomorphism ϕ : A →EndK(V ), that is, a representation of A.

Definition 2.31.

Let K be a field and A a K-algebra.
  1. (a)

    A representation of A over K is a K-vector space V  together with a K-algebra homomorphism θ : A →EndK(V ).

  2. (b)

    A matrix representation of A is a K-algebra homomorphism θ : AM n(K), for some n ≥ 1.

  3. (c)

    Suppose in (a) that V  is finite-dimensional. We may fix a basis in V  and write linear maps of V  as matrices with respect to such a basis. Then the representation of A becomes a K-algebra homomorphism θ : A → M n(K), that is, a matrix representation of A.


Example 2.32.

Let K be a field.
  1. (1)

    Let A be a K-subalgebra of M n(K), the algebra of n × n-matrices, then the inclusion map is an algebra homomorphism and hence is a matrix representation of A. Similarly, let A be a subalgebra of the K-algebra EndK(V ) of K-linear maps on V , where V  is a vector space over K. Then the inclusion map from A to EndK(V ) is an algebra homomorphism, hence it is a representation of A.

  2. (2)
    Consider the polynomial algebra K[X], and take a K-vector space V  together with a fixed linear transformation α. We have seen in Example 1.​24 that evaluation at α is an algebra homomorphism. Hence we have a representation of K[X] given by
$$\displaystyle \begin{aligned}\theta: K[X] \to \mathrm{End}_K(V), \ \ \theta(f) = f(\alpha). \end{aligned}$$
  3. (3)

    Let A = KG, the group algebra of a finite group. Define θ : A → M 1(K) = K by mapping each basis vector g ∈ G to 1, and extend to linear combinations. This is a representation of A: by Remark 2.6 it is enough to check the conditions on a basis. This is easy, since θ(g) = 1 for g ∈ G.


In Sect. 2.1 we introduced modules for an algebra as vector spaces on which the algebra acts by linear transformations. The following crucial result observes that modules and representations of an algebra are the same. Going from one notion to the other is a formal matter, nothing is ‘done’ to the modules or representations and it only describes two different views of the same thing.

Theorem 2.33.

Let K be a field and let A be a K-algebra.
  1. (a)
    Suppose V  is an A-module, with A-action A × V  V , (a, v)↦a  v. Then we have a representation of A given by
$$\displaystyle \begin{aligned}\theta: A\to \mathrm{End}_K(V)\mathit{\mbox{ , ~}}\theta(a)(v)= a\cdot v \mathit{\mbox{ for all }}a\in A, v\in V. \end{aligned}$$
  2. (b)
    Suppose θ : A →EndK(V ) is a representation of A. Then V  becomes an A-module by setting
$$\displaystyle \begin{aligned}A\times V\to V\mathit{\mbox{ , ~}}a\cdot v:= \theta(a)(v)\mathit{\mbox{ for all }}a\in A, v\in V. \end{aligned}$$

Roughly speaking, the representation θ corresponding to an A-module V  describes how each element a ∈ A acts linearly on the vector space V , and vice versa.


The proof basically consists of translating the module axioms from Definition 2.1 to the new language of representations from Definition 2.31, and vice versa.

(a) We first have to show that θ(a) ∈EndK(V ) for all a ∈ A. Recall from Lemma 2.5 that every A-module is also a K-vector space; then for all λ, μ ∈ K and v, w ∈ V  we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \theta(a)(\lambda v + \mu w) &\displaystyle = &\displaystyle \theta(a)(\lambda1_A\cdot v + \mu 1_A\cdot w) = (a\lambda 1_A)\cdot v + (a\mu 1_A)\cdot w \\ &\displaystyle = &\displaystyle \lambda (a\cdot v) + \mu (a\cdot w) = \lambda\theta(a)(v) + \mu\theta(a)(w), \end{array} \end{aligned} $$
where we again used axiom (Alg) from Definition 1.​1.
It remains to check that θ is an algebra homomorphism. First, it has to be a K-linear map; for any λ, μ ∈ K, a, b ∈ A and v ∈ V  we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \theta(\lambda a+\mu b)(v) &\displaystyle = &\displaystyle (\lambda a+\mu b)\cdot v = \lambda(a\cdot v) + \mu (b\cdot v) = (\lambda\theta(a)+\mu\theta(b))(v), \end{array} \end{aligned} $$
that is, θ(λa + μb) = λθ(a) + μθ(b). Next, for any a, b ∈ A and v ∈ V  we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} \theta(ab)(v) &\displaystyle = (ab)\cdot v = a\cdot( b\cdot v) = (\theta(a)\circ \theta(b))(v), \end{array} \end{aligned} $$
which holds for all v ∈ V , hence θ(ab) = θ(a) ∘ θ(b). Finally, it is immediate from the definition that θ(1A) = idV.

(b) This is analogous to the proof of part (a), where each argument can be reversed.□

Example 2.34.

Let K be a field.
  1. (1)

    Let V  be a K-vector space and A ⊆EndK(V ) a subalgebra. As observed in Example 2.32 the inclusion map θ : A →EndK(V ) is a representation. When V  is interpreted as an A-module we obtain precisely the ‘natural module’ from Example 2.4 with A-action given by A × V → V , (φ, v)↦φ(v).

  2. (2)

    The representation θ : K[X] →EndK(V ), θ(f) = f(α) for any K-linear map α on a K-vector space V  when interpreted as a K[X]-module is precisely the K[X]-module V α studied in Sect. 2.2.

  3. (3)
    Let A be a K-algebra, then A is an A-module with A-action given by the multiplication in A. The corresponding representation of A is then given by
$$\displaystyle \begin{aligned}\theta:A\to \mathrm{End}_K(A)\mbox{ , ~}\theta(a)(x)=ax \mbox{ for all }a,x\in A. \end{aligned}$$
    This representation θ is called the regular representation of A. (See also Exercise 1.​29.)

We now want to transfer the notion of module isomorphism (see Definition 2.20) to the language of representations. It will be convenient to have the following notion:

Definition 2.35.

Let K be a field. Suppose we are given two representations θ 1, θ 2 of a K-algebra A where θ 1 : A →EndK(V 1) and θ 2 : A →EndK(V 2). Then θ 1 and θ 2 are said to be equivalent if there is a vector space isomorphism ψ : V 1 → V 2 such that

$$\displaystyle \begin{aligned} \theta_1(a) = \psi^{-1}\circ \theta_2(a)\circ \psi \mbox{ ~for all }a\in A. \end{aligned} $$
$$\theta _1:A\to M_{n_1}(K)$$
$$\theta _2:A\to M_{n_2}(K)$$
are matrix representations of A, this means that θ 1(a) and θ 2(a) should be simultaneously similar, that is, there exists an invertible n 2 × n 2-matrix C (independent of a) such that θ 1(a) = C −1 θ 2(a)C for all a ∈ A. (Note that for equivalent representations the dimensions of the vector spaces must agree, that is, we have n 1 = n 2.)

For example, let A = K[X], and assume V 1, V 2 are finite-dimensional. Then (∗) holds for all a ∈ K[X] if and only if (∗) holds for a = X, that is, if and only if the matrices θ 1(X) and θ 2(X) are similar.

Proposition 2.36.

Let K be a field and A a K-algebra. Then two representations θ 1 : A →EndK(V 1) and θ 2 : A →EndK(V 2) of A are equivalent if and only if the corresponding A-modules V 1 and V 2 are isomorphic.


Suppose first that θ 1 and θ 2 are equivalent via the vector space isomorphism ψ : V 1 → V 2. We claim that ψ is also an A-module homomorphism (and then it is an isomorphism). By definition ψ : V 1 → V 2 is K-linear. Moreover, for v ∈ V 1 and a ∈ A we have

$$\displaystyle \begin{aligned}\psi(a\cdot v) = \psi(\theta_1(a)(v)) = \theta_2(a)(\psi(v)) = a\cdot \psi(v). \end{aligned}$$
Conversely, suppose ψ : V 1 → V 2 is an A-module isomorphism. Then for all a ∈ A and v ∈ V 1 we have

$$\displaystyle \begin{aligned}\psi(\theta_1(a)(v)) = \psi(a\cdot v) = a\cdot \psi(v)= \theta_2(a)(\psi(v)). \end{aligned}$$
This is true for all v ∈ V 1, that is, for all a ∈ A we have

$$\displaystyle \begin{aligned}\psi\circ \theta_1(a) = \theta_2(a)\circ \psi \end{aligned}$$
and hence the representations θ 1 and θ 2 are equivalent.□

Focussing on representations (rather than modules) gives some new insight. Indeed, an algebra homomorphism has a kernel which quite often is non-trivial. Factoring out the kernel, or an ideal contained in the kernel, gives a representation of a smaller algebra.

Lemma 2.37.

Let A be a K-algebra, and I a proper two-sided ideal of A, with factor algebra AI. Then the representations of AI are in bijection with the representations of A whose kernel contains I. The bijection takes a representation ψ of AI to the representation ψ  π where π : A  AI is the natural homomorphism.

Translating this to modules, there is a bijection between the set of AI-modules and the set of those A-modules V  for which xv = 0 for all x  I and v  V .


We prove the statement about the representations. The second statement then follows directly by Theorem 2.33.

(i) Let ψ : AI →EndK(V ) be a representation of the factor algebra AI, then the composition ψ ∘ π : A →EndK(V ) is an algebra homomorphism and hence a representation of A. By definition, I = ker(π) and hence I ⊆ker(ψ ∘ π).

(ii) Conversely, assume θ : A →EndK(V ) is a representation of A such that I ⊆ker(θ). By Theorem 1.​26, we have a well-defined algebra homomorphism 
$$\overline {\theta }: A/I\to \mathrm {End}_K(V)$$
defined by

$$\displaystyle \begin{aligned} \overline{\theta}(a+I) = \theta(a) \ \ (a\in A) \end{aligned}$$
and moreover 
$$\theta = \overline {\theta }\circ \pi $$
. Then 
$$\overline {\theta }$$
is a representation of AI.

In (i) we define a map between the two sets of representations, and in (ii) we show that it is surjective. Finally, let ψ, ψ be representations of AI such that ψ ∘ π = ψ ∘ π. The map π is surjective, and therefore ψ = ψ . Thus, the map in (i) is also injective.□

Remark 2.38.

In the above lemma, the AI-module V  is called inflation when it is viewed as an A-module. The two actions in this case are related by

$$\displaystyle \begin{aligned}(a+I)v = av \mbox{ ~for all }a\in A, v\in V. \end{aligned}$$

We want to illustrate the above inflation procedure with an example from linear algebra.

Example 2.39.

Let A = K[X] be the polynomial algebra. We take a representation θ : K[X] →EndK(V ) and let θ(X) =: α. The kernel is an ideal of K[X], so it is of the form ker(θ) = K[X]m = (m) for some polynomial m ∈ K[X]. Assume m ≠ 0, then we can take it to be monic. Then m is the minimal polynomial of α, as it is studied in linear algebra. That is, we have m(α) = 0 and f(α) = 0 if and only if f is a multiple of m in K[X].

By the above Lemma 2.37, the representation θ can be viewed as a representation of the algebra K[X]∕I whenever I is an ideal of K[X] contained in ker(θ) = (m), that is, I = (f) and f is a multiple of m.

2.5.1 Representations of Groups vs. Modules for Group Algebras

In this book we focus on representations of algebras. However, we explain in this section how the important theory of representations of groups can be interpreted in this context. Representations of groups have historically been the starting point for representation theory at the end of the 19th century. The idea of a representation of a group is analogous to that for algebras: one lets a group act by linear transformations on a vector space in a way compatible with the group structure. Group elements are invertible, therefore the linear transformations by which they act must also be invertible. The invertible linear transformations of a vector space form a group GL(V ), with respect to composition of maps, and it consists of the invertible elements in EndK(V ). For an n-dimensional vector space, if one fixes a basis, one gets the group GL n(K), which are the invertible matrices in M n(K), a group with respect to matrix multiplication.

Definition 2.40.

Let G be a group. A representation of G over a field K is a homomorphism of groups ρ : G → GL(V ), where V  is a vector space over K.

If V  is finite-dimensional one can choose a basis and obtain a matrix representation, that is, a group homomorphism ρ : G → GL n(K).

The next result explains that group representations are basically the same as representations for the corresponding group algebras (see Sect. 1.​1.​2). By Theorem 2.33 this can be rephrased by saying that a representation of a group is nothing but a module for the corresponding group algebra. Sometimes group homomorphisms are more useful, and sometimes it is useful that one can also use linear algebra.

Proposition 2.41.

Let G be a group and K a field.
  1. (a)

    Every representation ρ : G  GL(V ) of the group G over K extends to a representation θ ρ : KG →EndK(V ) of the group algebra KG given bygG α g g↦∑gG α g ρ(g).

  2. (b)

    Conversely, given a representation θ : KG →EndK(V ) of the group algebra KG, then the restriction ρ θ : G  GL(V ) of θ to G is a representation of the group G over K.



(a) Given a representation, that is, a group homomorphism ρ : G → GL(V ), we must show that θ ρ is an algebra homomorphism from KG to EndK(V ). First, θ ρ is linear, by definition, and maps into EndK(V ). Next, we must check that θ ρ(ab) = θ ρ(a)θ ρ(b). As mentioned in Remark 1.​23, it suffices to take a, b in a basis of KG, so we can take a, b ∈ G and then also ab ∈ G, and we have

$$\displaystyle \begin{aligned}\theta_{\rho}(ab) = \rho(ab) = \rho(a)\rho(b) = \theta_{\rho}(a)\theta_{\rho}(b). \end{aligned}$$
In addition θ ρ(1KG) = ρ(1G) = 1GL(V ) = idV.
(b) The elements of G form a subset (even a vector space basis) of the group algebra KG. Thus we can restrict θ to this subset and get a map

$$\displaystyle \begin{aligned}\rho_{\theta}:G \to GL(V)\mbox{ , ~}\rho_{\theta}(g)=\theta(g)\mbox{ for all } g\in G. \end{aligned}$$
In fact, every g ∈ G has an inverse g −1 in the group G; since θ is an algebra homomorphism it follows that

$$\displaystyle \begin{aligned}\theta(g)\theta(g^{-1}) = \theta(gg^{-1})=\theta(1_{KG})= \mathrm{id}_V, \end{aligned}$$
that is, ρ θ(g) = θ(g) ∈ GL(V ) is indeed an invertible linear map. Moreover, for every g, h ∈ G we have

$$\displaystyle \begin{aligned}\rho_{\theta}(gh)=\theta(gh) =\theta(g)\theta(h) =\rho_{\theta}(g)\rho_{\theta}(h) \end{aligned}$$
and ρ θ is a group homomorphism, as required.□

Example 2.42.

We consider a square in the plane, with corners (±1, ±1). Recall that the group of symmetries of the square is by definition the group of orthogonal transformations of 
$$\mathbb {R}^2$$
which leave the square invariant. This group is called dihedral group of order 8 and we denote it by D 4 (some texts call it D 8). It consists of four rotations (including the identity), and four reflections. Let r be the anti-clockwise rotation by π∕2, and let s be the reflection in the x-axis. One can check that D 4 = {s i r j | 0 ≤ i ≤ 1, 0 ≤ j ≤ 3}. We define this group as a subgroup of 
$$GL(\mathbb {R}^2)$$
. Therefore the inclusion map 
$$D_4\longrightarrow GL(\mathbb {R}^2) $$
is a group homomorphism, hence is a representation. We can write the elements in this group as matrices with respect to the standard basis. Then

$$\displaystyle \begin{aligned}\rho(r) = \left(\begin{matrix}0&-1 \\ 1&0\end{matrix}\right) \ \ \mbox{ and} \ \rho(s) = \left(\begin{matrix}1&0\\ 0&-1\end{matrix}\right). \end{aligned}$$
Then the above group homomorphism translates into a matrix representation

$$\displaystyle \begin{aligned}\rho:D_4\to GL_2(\mathbb{R})\mbox{ , ~}\rho(s^ir^j)= \rho(s)^i\rho(r)^j. \end{aligned}$$
By Proposition 2.41 this yields a representation θ ρ of the group algebra 
$$\mathbb {R}D_4$$
. Interpreting representations of algebras as modules (see Theorem 2.33) this means that 
$$V=\mathbb {R}^2$$
becomes an 
$$\mathbb {R} D_4$$
-module where each g ∈ D 4 acts on V  by applying the matrix ρ(g) to a vector.

Assume N is a normal subgroup of G with factor group GN, then it is reasonable to expect that representations of GN may be related to some representations of G. Indeed, we have the canonical map π : G → GN where π(g) = gN, which is a group homomorphism. In fact, using this we have an analogue of inflation for algebras, described in Lemma 2.37.

Lemma 2.43.

Let G be a group and N a normal subgroup of G with factor group GN, and consider representations over the field K. Then the representations of GN are in bijection with the representations of G whose kernel contains N. The bijection takes a representation θ of GN to the representation θ  π where π : G  GN is the canonical map.

Translating this to modules, this gives a bijection between the set of K(GN)-modules and the set of those KG-modules V  for which n  v = v for all n  N and v  V .


This is completely analogous to the proof of Lemma 2.37, so we leave it as an exercise. In fact, Exercise 2.24 shows that it is the same.□

2.5.2 Representations of Quivers vs. Modules for Path Algebras

The group algebra KG has basis the elements of the group G, which allows us to relate KG-modules and representations of G. The path algebra KQ of a quiver has basis the paths of Q. In analogy, we can define a representation of a quiver Q, which allows us to relate KQ-modules and representations of Q. We will introduce this now, and later we will study it in more detail.

Roughly speaking, a representation is as follows. A quiver consists of vertices and arrows, and if we want to realize it in the setting of vector spaces, we represent vertices by vector spaces, and arrows by linear maps, so that when arrows can be composed, the corresponding maps can also be composed.

Definition 2.44.

Let Q = (Q 0, Q 1) be a quiver. A representation 
$$\mathcal {V}$$
of Q over a field K is a set of K-vector spaces {V (i)∣i ∈ Q 0} together with K-linear maps V (α) : V (i) → V (j) for each arrow α from i to j. We sometimes also write 
$$\mathcal {V}$$
as a tuple 
$$\mathcal {V}=((V(i))_{i\in Q_0},(V(\alpha ))_{\alpha \in Q_1})$$

Example 2.45.

  1. (1)

    Let Q be the one-loop quiver as in Example 1.​13 with one vertex 1, and one arrow α with starting and end point 1. Then a representation 
$$\mathcal {V}$$
of Q consists of a K-vector space V (1) and a K-linear map V (α) : V (1) → V (1).

  2. (2)

    Let Q be the quiver 
$$1\, \stackrel {\alpha }{\longrightarrow }\, 2$$
. Then a representation 
$$\mathcal {V}$$
consists of two K-vector spaces V (1) and V (2) and a K-linear map V (α) : V (1) → V (2).

Consider the second example. We can construct from this a module for the path algebra KQ = span{e 1, e 2, α}. Take as the underlying space V := V (1) × V (2), a direct product of K-vector spaces (a special case of a direct product of modules as in Example 2.3 and Definition 2.17). Let e i act as the projection onto V (i) with kernel V (j) for j ≠ i. Then define the action of α on V  by

$$\displaystyle \begin{aligned} \alpha((v_1,v_2)):= V(\alpha)(v_1) \end{aligned}$$
(where v i ∈ V (i)). Conversely, if we have a KQ-module V  then we can turn it into a representation of Q by setting

$$\displaystyle \begin{aligned}V(1):= e_1V, \ V(2) = e_2V \end{aligned}$$
and V (α) : e 1 V → e 2 V  is given by (left) multiplication by α.

The following result shows how this can be done in general, it says that representations of a quiver Q over a field K are ‘the same’ as modules for the path algebra KQ.

Proposition 2.46.

Let K be a field and Q a quiver, with vertex set Q 0 and arrow set Q 1.
  1. (a)
$$\mathcal {V}=((V(i))_{i\in Q_0},(V(\alpha ))_{\alpha \in Q_1})$$
be a representation of Q over K. Then the direct product 
$$V:=\prod _{i\in Q_0} V(i)$$
becomes a KQ-module as follows: let 
$$v=(v_i)_{i\in Q_0}\in V$$
and let p = α rα 1 be a path in Q starting at vertex s(p) = s(α 1) and ending at vertex t(p) = t(α r). Then we set
$$\displaystyle \begin{aligned}p\cdot v = (0,\ldots,0,V(\alpha_r)\circ\ldots\circ V(\alpha_1)(v_{s(p)}),0,\ldots,0)\end{aligned}$$
    where the (possibly) non-zero entry is in position t(p). In particular, if r = 0, then e i ⋅ v = (0, …, 0, v i, 0, …, 0). This action is extended to all of KQ by linearity.
  2. (b)
    Let V  be a KQ-module. For any vertex i  Q 0 we set
$$\displaystyle \begin{aligned}V(i)=e_iV=\{e_i\cdot v\,|\,v\in V\}; \end{aligned}$$
    for any arrow 
$$i\stackrel {\alpha }{\longrightarrow } j$$
in Q 1 we set
$$\displaystyle \begin{aligned}V(\alpha):V(i)\to V(j)\mathit{\mbox{ , ~}}e_i\cdot v\mapsto \alpha\cdot(e_i\cdot v)=\alpha\cdot v. \end{aligned}$$
$$\mathcal {V}=((V(i))_{i\in Q_0},(V(\alpha ))_{\alpha \in Q_1})$$
is a representation of the quiver Q over K.
  3. (c)

    The constructions in (a) and (b) are inverse to each other.



(a) We check that the module axioms from Definition 2.1 are satisfied. Let p = α rα 1 and q be paths in Q. Since the KQ-action is defined on the basis of KQ and then extended by linearity, the distributivity (p + q) ⋅ v = p ⋅ v + q ⋅ v holds by definition. Moreover, let v, w ∈ V ; then

$$\displaystyle \begin{aligned} \begin{array}{rcl} p\cdot (v+w) &\displaystyle = &\displaystyle (0,\ldots,0,V(\alpha_r)\circ\ldots\circ V(\alpha_1)(v_{s(p)}+w_{s(p)}),0,\ldots,0) \\ &\displaystyle = &\displaystyle p\cdot v + p\cdot w \end{array} \end{aligned} $$
since all the maps V (α i) are K-linear. Since the multiplication in KQ is defined by concatenation of paths (see Sect. 1.​1.​3), it is immediate that p ⋅ (q ⋅ v) = (pq) ⋅ v for all v ∈ V  and all paths p, q in Q, and then by linearity also for arbitrary elements in KQ. Finally, the identity element is 
$$1_{KQ}=\sum _{i\in Q_0} e_i$$
, the sum of all trivial paths (see Sect. 1.​1.​3); by definition, e i acts by picking out the ith component, then for all v ∈ V  we have

$$\displaystyle \begin{aligned}1_{KQ}\cdot v = \sum_{i\in Q_0} e_i\cdot v = v. \end{aligned}$$
(b) According to Definition 2.44 we have to confirm that the V (i) = e i V  are K-vector spaces and that the maps V (α) are K-linear. The module axioms for V  imply that for every v, w ∈ V  and λ ∈ K we have

$$\displaystyle \begin{aligned}e_i\cdot v + e_i\cdot w = e_i\cdot (v+w) \in e_iV=V(i) \end{aligned}$$
and also

$$\displaystyle \begin{aligned}\lambda(e_i\cdot v) = (\lambda 1_{KQ}e_i)\cdot v = (e_i\lambda 1_{KQ})\cdot v = e_i\cdot (\lambda v) \in e_iV. \end{aligned}$$
So V (i) is a K-vector space for every i ∈ Q 0.
Finally, let us check that V (α) is a K-linear map for every arrow 
$$i\stackrel {\alpha }{\longrightarrow } j$$
in Q. Note first that αe i = α, so that the map V (α) is indeed given by e i ⋅ vα ⋅ v. Then for all λ, μ ∈ K and all v, w ∈ V (i) we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} V(\alpha)(\lambda v+\mu w) &\displaystyle = &\displaystyle \alpha\cdot (\lambda v+\mu w) = \alpha\cdot (\lambda 1_{KQ}\cdot v) + \alpha\cdot (\mu 1_{KQ}\cdot w) \\ &\displaystyle = &\displaystyle (\alpha\lambda 1_{KQ})\cdot v + (\alpha\mu 1_{KQ})\cdot w = (\lambda 1_{KQ}\alpha)\cdot v + (\mu 1_{KQ}\alpha)\cdot w \\ &\displaystyle = &\displaystyle \lambda 1_{KQ} \cdot (\alpha\cdot v) + \mu 1_{KQ}\cdot (\alpha\cdot w)= \lambda V(\alpha)(v) + \mu V(\alpha)(w). \end{array} \end{aligned} $$
So V (α) is a K-linear map.

(c) It is straightforward to check from the definitions that the two constructions are inverse to each other; we leave the details to the reader.□

Example 2.47.

We consider the quiver 
$$1\, \stackrel {\alpha }{\longrightarrow } \,2$$
. The 1-dimensional vector space span{e 2} is a KQ-module (more precisely, a KQ-submodule of the path algebra KQ). We interpret it as a representation of the quiver Q by 
$$0\, \stackrel {0}{\longrightarrow } \,\mathrm {span}\{e_2\}$$
. Also the two-dimensional vector space span{e 1, α} is a KQ-module. As a representation of Q this takes the form 
$$\mathrm {span}\{e_1\}\, \stackrel {V(\alpha )}{\longrightarrow } \,\mathrm {span}\{\alpha \}$$
, where V (α)(e 1) = α. Often, the vector spaces are only considered up to isomorphism, then the latter representation takes the more concise form 
$$K\, \stackrel {\mathrm {id}_K}{\longrightarrow } \,K$$


  1. 2.10.
    Let R be a ring. Suppose M is an R-module with submodules U, V  and W. We have seen in Exercise 2.3 that the sum U + V := {u + v | u ∈ U, v ∈ V } and the intersection U ∩ V  are submodules of M.
    1. (a)
      Show by means of an example that it is not in general the case that
$$\displaystyle \begin{aligned}U\cap (V+W) = (U\cap V) + (U\cap W). \end{aligned}$$
    2. (b)

      Show that U ∖ V  is never a submodule. Show also that the union U ∪ V  is a submodule if and only if U ⊆ V  or V ⊆ U.

  2. 2.11.
    Let A = M 2(K), the K-algebra of 2 × 2-matrices over a field K. Take the A-module M = A, and for i = 1, 2 define U i to be the subset of matrices where all entries not in the i-th column are zero. Moreover, let 
$$U_3:= \left \{ \left (\begin {matrix} a & a\\ b & b\end {matrix}\right )\mid a, b\in K\right \}$$
    1. (a)

      Check that each U i is an A-submodule of M.

    2. (b)

      Verify that for i ≠ j, the intersection U i ∩ U j is zero.

    3. (c)

      Show that M is not the direct sum of U 1, U 2 and U 3.

  3. 2.12.
    For a field K we consider the factor algebra A = K[X]∕(X 4 − 2). In each of the following cases find the number of 1-dimensional A-modules (up to isomorphism); moreover, describe explicitly the action of the coset of X on the module.
$$\displaystyle \begin{aligned} \text{(i) }K=\mathbb{Q}\text{; (ii) }K=\mathbb{R}\text{; (iii) }K=\mathbb{C}\text{; (iv) }K=\mathbb{Z}_3\text{; (v) }K=\mathbb{Z}_7. \end{aligned}$$
  4. 2.13.
$$A=\mathbb {Z}_p[X]/(X^n-1)$$
, where p is a prime number. We investigate 1-dimensional A-modules, that is, the roots of X n − 1 in 
$$\mathbb {Z}_p$$
(see Theorem 2.10).
    1. (a)

      Show that 
$$\bar {1}\in \mathbb {Z}_p$$
is always a root.

    2. (b)

      Show that the number of 1-dimensional A-modules is d where d is the greatest common divisor of n and p − 1. (Hint: You might use that the non-zero elements in 
$$\mathbb {Z}_p$$
are precisely the roots of the polynomial X p−1 − 1.)

  5. 2.14.
    For a field K we consider the natural T n(K)-module K n, where T n(K) is the algebra of upper triangular n × n-matrices.
    1. (a)
      In K n consider the K-subspaces
$$\displaystyle \begin{aligned}V_i:=\{(\lambda_1,\ldots,\lambda_n)^t\mid \lambda_j=0\mbox{ for all }j>i\}, \end{aligned}$$
      where 0 ≤ i ≤ n. Show that these are precisely the T n(K)-submodules of K n.
    2. (b)

      For any 0 ≤ j < i ≤ n consider the factor module V i,j := V iV j. Show that these give 
$$\frac {n(n+1)}{2}$$
pairwise non-isomorphic T n(K)-modules.

    3. (c)

      For each standard basis vector e i ∈ K n, where 1 ≤ i ≤ n, determine the annihilator 
$$\mathrm {Ann}_{T_n(K)}(e_i)$$
(see Example 2.25) and identify the factor module 
$$T_n(K)/\mathrm {Ann}_{T_n(K)}(e_i)$$
with a T n(K)-submodule of K n, up to isomorphism.

  6. 2.15.

    Let R be a ring and suppose M = U × V , the direct product of R-modules U and V . Check that 
$$\tilde {U}:= \{ (u, 0)\mid u\in U\}$$
is a submodule of M, isomorphic to U. Write down a similar submodule 
$$\tilde {V}$$
of M isomorphic to V , and show that 
$$M= \tilde {U} \oplus \tilde {V}$$
, the direct sum of submodules.

  7. 2.16.
    Let K be a field. Assume M, M 1, M 2 are K-vector spaces and α i : M i → M are K-linear maps. The pull-back of (α 1, α 2) is defined to be
$$\displaystyle \begin{aligned}E:= \{ (m_1, m_2)\in M_1\times M_2\mid \alpha_1(m_1) + \alpha_2(m_2) =0\}.\end{aligned} $$
    1. (a)

      Check that E is a K-subspace of M 1 × M 2.

    2. (b)

      Assume that M, M 1, M 2 are finite-dimensional K-vector spaces and that M = im(α 1) + im(α 2). Show that then for the vector space dimensions we have dimK E =dimK M 1 +dimK M 2 −dimK M. (Hint: Show that the map (m 1, m 2)↦α 1(m 1) + α 2(m 2) from M 1 × M 2 to M is surjective, and has kernel E.)

    3. (c)

      Now assume that M, M 1, M 2 are A-modules where A is some K-algebra, and that α 1 and α 2 are A-module homomorphisms. Show that then E is an A-submodule of M 1 × M 2.

  8. 2.17.
    Let K be a field. Assume W, M 1, M 2 are K-vector spaces and β i : W → M i are K-linear maps. The push-out of (β 1, β 2) is defined to be
$$\displaystyle \begin{aligned} F:= (M_1\times M_2)/ C, \end{aligned}$$
    where C := {(β 1(w), β 2(w))∣w ∈ W}.
    1. (a)

      Check that C is a K-subspace of M 1 × M 2, hence F is a K-vector space.

    2. (b)
      Assume W, M 1, M 2 are finite-dimensional and ker(β 1) ∩ker(β 2) = 0. Show that then
$$\displaystyle \begin{aligned}\dim_K F = \dim_K M_1 + \dim_K M_2 - \dim_K W. \end{aligned}$$
      (Hint: Show that the linear map w↦(β 1(w), β 2(w)) from W to C is an isomorphism.)
    3. (c)

      Assume that W, M 1, M 2 are A-modules where A is some K-algebra and that β 1, β 2 are A-module homomorphisms. Show that then C and hence F are A-modules.

  9. 2.18.
    Let E be the pull-back as in Exercise 2.16 and assume M = im(α 1) + im(α 2). Now take the push-out F as in Exercise 2.17 where W = E with the same M 1, M 2, and where β i : W → M i are the maps
$$\displaystyle \begin{aligned}\beta_1(m_1, m_2) := m_1, \ \beta_2(m_1, m_2) := m_2 \ \ \ (\mbox{where }(m_1, m_2)\in W=E). \end{aligned} $$
    1. (a)

      Check that ker(β 1) ∩ker(β 2) = 0.

    2. (b)

      Show that the vector space C in the construction of the pushout is equal to E.

    3. (c)

      Show that the pushout F is isomorphic to M. (Hint: Consider the map from M 1 × M 2 to M defined by (m 1, m 2)↦α 1(m 1) + α 2(m 2).)

  10. 2.19.

    Let K be a field and KG be the group algebra where G is a finite group. Recall from Example 2.4 that the trivial KG-module is the 1-dimensional module with action g ⋅ x = x for all g ∈ G and x ∈ K, linearly extended to all of KG. Show that the corresponding representation θ : KG →EndK(K) satisfies θ(a) = idK for all a ∈ KG. Check that this is indeed a representation.

  11. 2.20.
    Let G be the group of symmetries of a regular pentagon, that is, the group of orthogonal transformations of 
$$\mathbb {R}^2$$
which leave the pentagon invariant. That is, G is the dihedral group of order 10, a subgroup of 
$$GL(\mathbb {R}^2)$$
. As a group, G is generated by the (counterclockwise) rotation by 
$$\frac {2\pi i}{5}$$
, which we call r, and a reflection s; the defining relations are 
$$r^5=\mathrm {id}_{\mathbb {R}^2}$$
$$s^2=\mathrm {id}_{\mathbb {R}^2}$$
and s −1 rs = r −1. Consider the group algebra 
$$\mathbb {C}G$$
of G over the complex numbers, and suppose that 
$$\omega \in \mathbb {C}$$
is some 5-th root of unity. Show that the matrices
$$\displaystyle \begin{aligned}\rho(r) = \left(\begin{matrix} \omega &amp;0\\ 0&amp;\omega^{-1} \end{matrix}\right), \ \ \rho(s) = \left(\begin{matrix}0&amp;1\\ 1&amp;0\end{matrix}\right) \end{aligned}$$
    satisfy the defining relations for G, hence give rise to a group representation 
$$\rho : G \to GL_2(\mathbb {C})$$
, and a 2-dimensional 
$$\mathbb {C}G$$
  12. 2.21.
    Let K be a field. We consider the quiver Q given by 
$$1 \stackrel {\alpha }{\longleftarrow } 2 \stackrel {\beta }{\longleftarrow } 3 $$
and the path algebra KQ as a KQ-module.
    1. (a)

      Let V := span{e 2, α}⊆ KQ. Explain why V = KQe 2 and hence V  is a KQ-submodule of KQ.

    2. (b)

      Find a K-basis of the KQ-submodule W := KQβ generated by β.

    3. (c)

      Express the KQ-modules V  and W as a representation of the quiver Q. Are V  and W isomorphic as KQ-modules?

  13. 2.22.
    Let A = KQ where Q is the following quiver: 
$$ 1 \stackrel {\alpha }\longrightarrow 2 \stackrel {\beta }\longleftarrow 3$$
. This exercise illustrates that A as a left module and A as a right module have different properties.
    1. (a)

      As a left module A = Ae 1 ⊕ Ae 2 ⊕ Ae 3 (see Exercise 2.6). For each Ae i, find a K-basis, and verify that each of Ae 1 and Ae 3 is 2-dimensional, and Ae 2 is 1-dimensional.

    2. (b)

      Show that the only 1-dimensional A-submodule of Ae 1 is span{α}. Deduce that Ae 1 cannot be expressed as Ae 1 = U ⊕ V  where U and V  are non-zero A-submodules.

    3. (c)

      Explain briefly why the same holds for Ae 3.

    4. (d)

      As a right A-module, A = e 1 A ⊕ e 2 A ⊕ e 3 A (by the same reasoning as in Exercise 2.6). Verify that e 1 A and e 3 A are 1-dimensional.

  14. 2.23.
    Assume A = K[X] and let f = gh where g and h are polynomials in A. Then Af = (f) ⊆ (g) = Ag and the factor module is
$$\displaystyle \begin{aligned}(g)/(f) = \{ r g + (f)\mid r\in K[X] \}. \end{aligned}$$
    Show that (g)∕(f) is isomorphic to K[X]∕(h) as a K[X]-module.
  15. 2.24.
    Let G be a group and N a normal subgroup of G, and let A = KG, the group algebra over the field K.
    1. (a)

      Show that the space 
$$I=\mathrm {span}\{ g_1-g_2\mid g_1g_2^{-1} \in N\} $$
is a 2-sided ideal of A.

    2. (b)

      Show that the A-modules on which I acts as zero are precisely the A-modules V  such that n ⋅ v = v for all n ∈ N and v ∈ V .

    3. (c)

      Explain briefly how this connects Lemmas 2.37 and 2.43.