1.2 Subalgebras, Ideals and Factor Algebras

In analogy to the ‘subspaces’ of vector spaces, and ‘subgroups’ of groups, we should define the notion of a ‘subalgebra’. Suppose $A$ is a $K$-algebra, then, roughly speaking, a subalgebra $B$ of $A$ is a subset of $A$ which is itself an algebra with respect to the operations in $A$. This is made precise in the following definition.

Definition 1.14.

Let $K$ be a field and $A$ a $K$-algebra. A subset $B$ of $A$ is called a K-subalgebra (or just subalgebra) of $A$ if the following holds:
  1. (i)

    $B$ is a $K$-subspace of $A$, that is, for every $λ,μ∈K$ and $b_1,b_2∈B$ we have $λb_1+μb_2∈B$.

  2. (ii)

    $B$ is closed under multiplication, that is, for all $b_1$, $b_2∈B$, the product $b_1b_2$ belongs to $B$.

  3. (iii)

    The identity element 1$A$ of $A$ belongs to $B$.


Exercise 1.3.

Let $A$ be a $K$-algebra and $B⊆A$ a subset. Show that $B$ is a $K$-subalgebra of $A$ if and only if $B$ itself is a $K$-algebra with the operations induced from $A$.

Remark 1.15.

Suppose $B$ is given as a subset of some algebra $A$, with the same addition, scalar multiplication, and multiplication. To decide whether or not $B$ is an algebra, there is no need to check all the axioms of Definition 1.1. Instead, it is enough to verify conditions (i) to (iii) of Definition 1.14.

We consider several examples.

Example 1.16.

Let $K$ be a field.
  1. (1)
    The $K$-algebra $M_n(K)$ of $n×n$-matrices over $K$ has many important subalgebras.
    1. (i)
      The upper triangular matrices \begin{aligned}T_n(K):= \{ a=(a_{ij})∈ M_n(K)∣ a_{ij}=0\mbox{ for }i>j\} \end{aligned} form a subalgebra of $M_n(K)$. Similarly, one can define lower triangular matrices and they also form a subalgebra of $M_n(K)$.
    2. (ii)
      The diagonal matrices \begin{aligned}D_n(K):= \{ a=(a_{ij})∈ M_n(K)∣ a_{ij}=0\mbox{ for }i≠ j\} \end{aligned} form a subalgebra of $M_n(K)$.
    3. (iii)
      The three-subspace algebra is the subalgebra of $M_4(K)$ given by
      \begin{aligned}\left\{ \begin{pmatrix} a_1&b_1&b_2&b_3 \\ 0&a_2&0&0 \\ 0&0&a_3&0 \\ 0&0&0&a_4\end{pmatrix}∣ a_i, b_j ∈ K \right\}. \end{aligned}
    4. (iv)
      There are also subalgebras such as
      \begin{aligned}\left\{ \begin{pmatrix} a&b&0&0 \\ c&d&0&0 \\ 0&0&x&y\\ 0&0&z&u\end{pmatrix}∣ a, b, c, d, x, y, z, u ∈ K \right\} ⊆ M_4(K). \end{aligned}
  2. (2)

    The $n×n$-matrices $M_n(ℤ)⊆ M_n(ℝ)$ over the integers are closed under addition and multiplication, but $M_n(ℤ)$ is not an $ℝ$-subalgebra of $M_n(ℝ)$ since it is not an $ℝ$-subspace.

  3. (3)
    The subset \begin{aligned}\left\{ \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix}∣ b∈ K\right\}⊆ T_2(K) \end{aligned} is not a $K$-subalgebra of $T_2(K)$ since it does not contain the identity element.
  4. (4)

    Let $A$ be a $K$-algebra. For any element $a∈A$ define $A_a$ to be the $K$-span of {1$A$, $a,a^2$, …}. That is, $A_a$ is the space of polynomial expressions in $a$. This is a $K$-subalgebra of $A$, and it is always commutative. Note that if $A$ is finite-dimensional then so is $A_a$.

  5. (5)

    Let $A=A_1×A_2$, the direct product of two algebras. Then $A_1$ ×{0} is not a subalgebra of $A$ since it does not contain the identity element of $A$.

  6. (6)

    Let $H$ be a subgroup of a group $G$. Then the group algebra $KH$ is a subalgebra of the group algebra $KG$.

  7. (7)

    Let $KQ$ be the path algebra of the quiver $1 \xleftarrow{α} 2 \xleftarrow{β} 3$. We can consider the ‘subquiver’, $Q'$ given by $1 \xleftarrow{α} 2$. The path algebra $KQ'$ is not a subalgebra of $KQ$ since it does not contain the identity element $1_{KQ}=e_1+e_2+e_3$ of $KQ$.


Exercise 1.4.

Verify that the three-subspace algebra from Example 1.16 is a subalgebra of $M_4(K)$, hence is a $K$-algebra.

In addition to subalgebras, there are also ideals, and they are needed when one wants to define factor algebras.

Definition 1.17.

If $A$ is a $K$-algebra then a subset $I$ is a left ideal of $A$ provided $(I,+)$ is a subgroup of $(A,+)$ such that $ax∈I$ for all $x∈I$ and $a∈A$. Similarly, $I$ is a right ideal of $A$ if $(I,+)$ is a subgroup of $(A,+)$ such that $xa∈I$ for all $x∈I$ and $a∈A$. A subset $I$ of $A$ is a two-sided ideal (or just ideal) of $A$ if it is both a left ideal and a right ideal.

Remark 1.18.

  1. (1)

    The above definition works verbatim for rings instead of algebras.

  2. (2)

    For commutative algebras $A$ the notions of left ideal, right ideal and two-sided ideal clearly coincide. However, for non-commutative algebras, a left ideal might not be a right ideal, and vice versa; see the examples below.

  3. (3)

    An ideal $I$ (left or right or two-sided) of an algebra $A$ is by definition closed under multiplication, and also a subspace, see Lemma 1.20 below. However, an ideal is in general not a subalgebra, since the identity element need not be in $I$. Actually, a (left or right or two-sided) ideal $I⊆A$ contains 1$A$ if and only if $I=A$. In addition, a subalgebra is in general not an ideal.


Exercise 1.5.

Assume $B$ is a subalgebra of $A$. Show that $B$ is a left ideal (or right ideal) if and only if $B=A$.

Example 1.19.

  1. (1)

    Every $K$-algebra $A$ has the trivial two-sided ideals {0} and $A$. In the sequel, we will usually just write 0 for the ideal {0}.

  2. (2)

    Let $A$ be a $K$-algebra. For every $z∈A$ the subset $Az$ = {$Az$∣$a∈A$} is a left ideal of $A$. Similarly, $za$ = {$za$∣$a∈A$} is a right ideal of $A$ for every $z∈A$. These are called the principal (left or right) ideals generated by $z$. As a notation for principal ideals in commutative algebras we often use $Az=(z)$.

  3. (3)

    For non-commutative algebras, $Az$ need not be a two-sided ideal. For instance, let $A=M_n(K)$ for some $n$ ≥ 2 and consider the matrix unit $z=E_{ii}$ for some $i$ ∈{1, …, $n$}. Then the left ideal $Az$ consists of the matrices with non-zero entries only in column $i$, whereas the right ideal $zA$ consists of those matrices with non-zero entries in row $i$. In particular, $Az≠zA$, and both are not two-sided ideals of $M_n(K)$.

  4. (4)
    The only two-sided ideals in $M_n(K)$ are the trivial ideals. In fact, suppose that $I≠0$ is a two-sided ideal in $M_n(K)$, and let $a=(a_{ij}$)$∈I$ ∖{0}, say $a_{kℓ}≠0$. Then for every $r,s$ ∈{1, …, $n$} we have that $E_{rk}aE_{ℓs}∈I$ since $I$ is a two-sided ideal. On the other hand, \begin{aligned} E_{rk}aE_{ℓ s} &= E_{rk} \left( \sum_{i,j=1}^n a_{ij}E_{ij}\right) E_{ℓ s} = \left( \sum_{j=1}^n a_{kj}E_{rj}\right) E_{ℓ s} = a_{kℓ} E_{rs}. \end{aligned} Since $a_{kℓ}≠0$ we conclude that $E_{rs}∈I$ for all $r,s$ and hence $I=M_n(K)$.
  5. (5)

    Consider the $K$-algebra $T_n(K)$ of upper triangular matrices. The $K$-subspace of strict upper triangular matrices $I_1$ := span{$E_{ij}$ | $1≤i < j≤n$} forms a two-sided ideal. More generally, for any $d∈ ℕ$ the subspace $I_d$ := span{$E_{ij}$ | $d≤j−i$} is a two-sided ideal of $T_n(K)$. Note that by definition, $I_d$=0 for $d≥n$.

  6. (6)

    Let $Q$ be a quiver and $A=KQ$ be its path algebra over a field $K$. Take the trivial path $e_i$ for a vertex $i∈Q_0$. Then by Example (2) above we have the left ideal Ae$i$. This is spanned by all paths in $Q$ with starting vertex $i$. Similarly, the right ideal $e_iA$ is spanned by all paths in $Q$ ending at vertex $i$.

    A two-sided ideal of $KQ$ is for instance given by the subspace ($KQ$)≥1 spanned by all paths in $Q$ of non-zero length (that is, by all paths in $Q$ except the trivial paths). More generally, for every $d∈ ℕ$ the linear combinations of all paths in $Q$ of length at least $d$ form a two-sided ideal ($KQ$)≥$d$.


Exercise 1.6.

Let $Q$ be a quiver, $i$ a vertex in $Q$ and $A=KQ$ the path algebra (over some field $K)$. Find a condition on $Q$ so that the left ideal Ae$i$ is a two-sided ideal.

Let $A$ be a $K$-algebra and let $I⊂A$ be a proper two-sided ideal. Recall from basic algebra that the cosets $A/I$ := {$a+I$ | $a∈A$} form a ring, the factor ring, with addition and multiplication defined by \begin{aligned}(a+I)+(b+I):= a+b+I, \ \ \ (a+I)(b+I):= ab+I \end{aligned} for $a,b∈A$. Note that these operations are well-defined, that is, they are independent of the choice of the representatives of the cosets, because $I$ is a two-sided ideal. Moreover, the assumption $I≠A$ is needed to ensure that the factor ring has an identity element; see Axiom (R7) in Sect. 1.1.

For $K$-algebras we have some extra structure on the factor rings.

Lemma 1.20.

Let A be a K-algebra. Then the following holds.
  1. (a)

    Every left (or right) ideal I of A is a K-subspace of A.

  2. (b)

    If I is a proper two-sided ideal of A then the factor ring AI is a K-algebra, the factor algebra of A with respect to I.



(a) Let $I$ be a left ideal. By definition, $(I,+)$ is an abelian group. We need to show that if $λ∈K$ and $x∈I$ then $λx∈I$. But $λ$1$A$$∈A$, and we obtain \begin{aligned}λ x= λ(1_Ax) = (λ 1_A)x ∈ I, \end{aligned} since $I$ is a left ideal. The same argument works if $I$ is a right ideal, by axiom (Alg) in Definition 1.1.
(b) We have already recalled above that the cosets $A/I$ form a ring. Moreover, by part (a), $I$ is a $K$-subspace and hence $A/I$ is also a $K$-vector space with (well-defined) scalar multiplication $λ(a+I)=λa+I$ for all $λ∈K$ and $a∈A$. According to Definition 1.1 it only remains to show that axiom (Alg) holds. But this property is inherited from $A$; explicitly, let $λ∈K$ and $a,b∈A$, then \begin{aligned} λ((a+I)(b+I)) &= λ(ab + I) = λ(ab) + I = (λ a)b+I \\ &= (λ a+I)(b+I) = (λ (a+I))(b+I). \end{aligned} Similarly, using that $λ$(ab)$=a$(λb) by axiom (Alg), one shows that $λ((a+I)(b+I)$) = $(a+I)(λ(b+I)$).□

Example 1.21.

Consider the algebra $K[X]$ of polynomials in one variable over a field $K$. Recall from a course on basic algebra that every non-zero ideal $I$ of $K[X]$ is of the form $K[X]f=(f)$ for some non-zero polynomial $f∈K[X]$ (that is, $K[X]$ is a principal ideal domain). The factor algebra $A/I=K[X]/(f)$ is finite-dimensional. More precisely, we claim that it has dimension equal to the degree $d$ of the polynomial $f$ and that a $K$-basis of $K[X]/(f)$ is given by the cosets $1+(f)$, $X+(f)$, …, $X$$d$−1$+(f)$. In fact, if $g∈K[X]$ then division with remainder in $K[X]$ (polynomial long division) yields $g=qf+r$ with polynomials $q,r∈K[X]$ and $r$ has degree less than $d$, the degree of $f$. Hence \begin{aligned}g+(f) = r+(f)∈ \mathrm{span}\{1+(f),X+(f),…,X^{d-1}+(f)\}. \end{aligned} On the other hand, considering degrees one checks that this spanning set of $K[X]/(f)$ is also linearly independent.