# 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 ${\displaystyle F}$ 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 porduct 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