# Abstract Algebra/Algebras

In this section we will talk about structures with three operations. These are called algebras. We will start by defining an algebra over a field, which is a vector space with a bilinear vector product. After giving some examples, we will then move to a discussion of quivers and their path algebras.

## Algebras over a Field

Definition 1: Let ${\displaystyle F}$  be a field, and let ${\displaystyle A}$  be an ${\displaystyle F}$ -vector space on which we define the vector product ${\displaystyle \cdot \,:\,A\times A\rightarrow A}$ . Then ${\displaystyle A}$  is called an algebra over ${\displaystyle F}$  provided that ${\displaystyle (A,+,\cdot )}$  is a ring, where ${\displaystyle +}$  is the vector space addition, and if for all ${\displaystyle a,b,c\in A}$  and ${\displaystyle \alpha \in F}$ ,

1. ${\displaystyle a(bc)=(ab)c}$ ,
2. ${\displaystyle a(b+c)=ab+ac}$  and ${\displaystyle (a+b)c=ac+bc}$ ,
3. ${\displaystyle \alpha (ab)=(\alpha a)b=a(\alpha b)}$ .

The dimension of an algebra is the dimension of ${\displaystyle A}$  as a vector space.

Remark 2: The appropriate definition of a subalgebra is clear from Definition 1. We leave its formal statement to the reader.

Definition 2: If ${\displaystyle (A,+,\cdot )}$  is a commutative ring, ${\displaystyle A}$  is called a commutative algebra. If it is a division ring, ${\displaystyle A}$  is called a division algebra. We reserve the terms real and complex algebra for algebras over ${\displaystyle \mathbb {R} }$  and ${\displaystyle \mathbb {C} }$ , respectively.

The reader is invited to check that the following examples really are examples of algebras.

Example 3: Let ${\displaystyle F}$  be a field. The vector space ${\displaystyle F^{n}}$  forms a commutative ${\displaystyle F}$ -algebra under componentwise multiplication.

Example 4: The quaternions ${\displaystyle \mathbb {H} }$  is a 4-dimensional real algebra. We leave it to the reader to show that it is not a 2-dimensional complex algebra.

Example 5: Given a field ${\displaystyle F}$ , the vector space of polynomials ${\displaystyle F[x]}$  is a commutative ${\displaystyle F}$ -algebra in a natural way.

Example 6: Let ${\displaystyle F}$  be a field. Then any matrix ring over ${\displaystyle F}$ , for example ${\displaystyle \left({\begin{array}{cc}F&0\\F&F\end{array}}\right)}$ , gives rise to an ${\displaystyle F}$ -algebra in a natural way.

## Quivers and Path Algebras

Naively, a quiver can be understood as a directed graph where we allow loops and parallell edges. Formally, we have the following.

Definition 7: A quiver is a collection of four pieces of data, ${\displaystyle Q=(Q_{0},Q_{1},s,t)}$ ,

1. ${\displaystyle Q_{0}}$  is the set of vertices of the quiver,
2. ${\displaystyle Q_{1}}$  is the set of edges, and
3. ${\displaystyle s,t\,:\,Q_{1}\rightarrow Q_{0}}$  are functions associating with each edge a source vertex and a target vertex, respectively.

We will always assume that ${\displaystyle Q_{0}}$  is nonempty and that ${\displaystyle Q_{0}}$  and ${\displaystyle Q_{1}}$  are finite sets.

Example 8: The following are the simplest examples of quivers:

1. The quiver with one point and no edges, represented by ${\displaystyle 1}$ .
2. The quiver with ${\displaystyle n}$  point and no edges, ${\displaystyle 1\quad 2\quad ...\quad n}$ .
3. The linear quiver with ${\displaystyle n}$  points, ${\displaystyle 1\,{\stackrel {a_{1}}{\longrightarrow }}\,2\,{\stackrel {a_{2}}{\longrightarrow }}\,...\,{\xrightarrow {a_{n-1}}}\,n}$ .
4. The simplest quiver with a nontrivial loop, ${\displaystyle 1{\underset {a}{\stackrel {b}{\leftrightarrows }}}2}$ .

Definition 9: Let ${\displaystyle Q}$  be a quiver. A path in ${\displaystyle Q}$  is a sequence of edges ${\displaystyle a=a_{m}a_{m-1}...a_{1}}$  where ${\displaystyle s(a_{i})=t(a_{i-1})}$  for all ${\displaystyle i=2,...,m}$ . We extend the domains of ${\displaystyle s}$  and ${\displaystyle t}$  and define ${\displaystyle s(a)\equiv s(a_{0})}$  and ${\displaystyle t(a)\equiv t(a_{m})}$ . We define the length of the path to be the number of edges it contains and write ${\displaystyle l(a)=m}$ . With each vertex ${\displaystyle i}$  of a quiver we associate the trivial path ${\displaystyle e_{i}}$  with ${\displaystyle s(e_{i})=t(e_{i})=i}$  and ${\displaystyle l(e_{i})=0}$ . A nontrivial path ${\displaystyle a}$  with ${\displaystyle s(a)=t(a)=i}$  is called an oriented loop at ${\displaystyle i}$ .

The reason quivers are interesting for us is that they provide a concrete way of constructing a certain family of algebras, called path algebras.

Definition 10: Let ${\displaystyle Q}$  be a quiver and ${\displaystyle F}$  a field. Let ${\displaystyle FQ}$  denote the free vector space generated by all the paths of ${\displaystyle Q}$ . On this vector space, we define a vector product in the obvious way: if ${\displaystyle u=u_{m}...u_{1}}$  and ${\displaystyle v=v_{n}...v_{1}}$  are paths with ${\displaystyle s(v)=t(u)}$ , define their product ${\displaystyle vu}$  by concatenation: ${\displaystyle vu=v_{n}...v_{1}u_{m}...u_{1}}$ . If ${\displaystyle s(v)\neq t(u)}$ , define their product to be ${\displaystyle vu=0}$ . This product turns ${\displaystyle FQ}$  into an ${\displaystyle F}$ -algebra, called the path algebra of ${\displaystyle Q}$ .

Lemma 11: Let ${\displaystyle Q}$  be a quiver and ${\displaystyle F}$  field. If ${\displaystyle Q}$  contains a path of length ${\displaystyle |Q_{0}|}$ , then ${\displaystyle FQ}$  is infinite dimensional.

Proof: By a counting argument such a path must contain an oriented loop, ${\displaystyle a}$ , say. Evidently ${\displaystyle \{a^{n}\}_{n\in \mathbb {N} }}$  is a linearly independent set, such that ${\displaystyle FQ}$  is infinite dimensional.

Lemma 12: Let ${\displaystyle Q}$  be a quiver and ${\displaystyle F}$  a field. Then ${\displaystyle FQ}$  is infinite dimensional if and only if ${\displaystyle Q}$  contains an oriented loop.

Proof: Let ${\displaystyle a}$  be an oriented loop in ${\displaystyle Q}$ . Then ${\displaystyle FQ}$  is infinite dimensional by the above argument. Conversely, assume ${\displaystyle Q}$  has no loops. Then the vertices of the quiver can be ordered such that edges always go from a lower to a higher vertex, and since the length of any given path is bounded above by ${\displaystyle |Q_{0}|-1}$ , there dimension of ${\displaystyle FQ}$  is bounded above by ${\displaystyle \mathrm {dim} \,FQ\leq |Q_{0}|^{2}-|Q_{0}|<\infty }$ .

Lemma 13: Let ${\displaystyle Q}$  be a quiver and ${\displaystyle F}$  a field. Then the trivial edges ${\displaystyle e_{i}}$  form an orthogonal idempotent set.

Proof: This is immediate from the definitions: ${\displaystyle e_{i}e_{j}=0}$  if ${\displaystyle i\neq j}$  and ${\displaystyle e_{i}^{2}=e_{i}}$ .

Corollary 14: The element ${\displaystyle \sum _{i\in Q_{0}}e_{i}}$  is the identity element in ${\displaystyle FQ}$ .

Proof: It sufficed to show this on the generators of ${\displaystyle FQ}$ . Let ${\displaystyle a}$  be a path in ${\displaystyle Q}$  with ${\displaystyle s(a)=j}$  and ${\displaystyle t(a)=k}$ . Then ${\displaystyle \left(\sum _{i\in Q_{0}}e_{i}\right)a=\sum _{i\in Q_{0}}e_{i}a=e_{j}a=a}$ . Similarily, ${\displaystyle a\left(\sum _{i\in Q_{0}}e_{i}\right)=a}$ .

To be covered:

- General R-algebras