**1.1**Definition and Examples**1.2**Subalgebras, Ideals and Factor Algebras**1.3**Algebra Homomorphisms**1.4**Some Algebras of Small Dimensions

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.

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:

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

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

- (iii)
The identity element 1

_{$A$}of $A$ belongs to $B$.

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 $A$.

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.

Let $K$ be a field.

- (1)The $K$-algebra $M_n(K)$ of $n×n$-matrices over $K$ has many important subalgebras.
- (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 $M_n(K)$.
- (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 $M_n(K)$.
- (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} - (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}

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

- (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 $KQ$.

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.

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.

- (1)
The above definition works verbatim for rings instead of algebras.

- (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)
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.

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

- (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)
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)
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)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)
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)
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 ($KQ$)^{≥$d$}.

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.

- (a)
*Every left (or right) ideal I of A is a K-subspace of A.* - (b)
*If I is a proper two-sided ideal of A then the factor ring A*∕*I 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 $λ((a+I)(b+I)$) = $(a+I)(λ(b+I)$).□

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.