# 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 FieldEdit

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

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

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

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

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

Example 4: The quaternions $\mathbb{H}$ 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 $F$, the vector space of polynomials $F[x]$ is a commutative $F$-algebra in a natural way.

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

## Quivers and Path AlgebrasEdit

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, $Q=(Q_0,Q_1,s,t)$,

1. $Q_0$ is the set of vertices of the quiver,
2. $Q_1$ is the set of edges, and
3. $s,t\,:\, Q_1\rightarrow Q_0$ are functions associating with each edge a source vertex and a target vertex, respectively.

We will always assume that $Q_0$ is nonempty and that $Q_0$ and $Q_1$ are finite sets.

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

1. The quiver with one point and no edges, represented by $1$.
2. The quiver with $n$ point and no edges, $1\quad 2\quad ... \quad n$.
3. The linear quiver with $n$ points, $1\,\stackrel{a_1}{\longrightarrow}\, 2\,\stackrel{a_2}{\longrightarrow} \,...\,\xrightarrow{a_{n-1}} \,n$.
4. The simplest quiver with a nontrivial loop, $1\underset{a}\stackrel{b}{\leftrightarrows} 2$.

Definition 9: Let $Q$ be a quiver. A path in $Q$ is a sequence of edges $a=a_ma_{m-1}...a_1$ where $s(a_i)=t(a_{i-1})$ for all $i=2,...,m$. We extend the domains of $s$ and $t$ and define $s(a)\equiv s(a_0)$ and $t(a)\equiv t(a_m)$. We define the length of the path to be the number of edges it contains and write $l(a)=m$. With each vertex $i$ of a quiver we associate the trivial path $e_i$ with $s(e_i)=t(e_i)=i$ and $l(e_i)=0$. A nontrivial path $a$ with $s(a)=t(a)=i$ is called an oriented loop at $i$.

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 $Q$ be a quiver and $F$ a field. Let $FQ$ denote the free vector space generated by all the paths of $Q$. On this vector space, we define a vector porduct in the obvious way: if $u=u_m...u_1$ and $v=v_n...v_1$ are paths with $s(v)=t(u)$, define their product $vu$ by concatenation: $vu=v_n...v_1u_m...u_1$. If $s(v)\neq t(u)$, define their product to be $vu=0$. This product turns $FQ$ into an $F$-algebra, called the path algebra of $Q$.

Lemma 11: Let $Q$ be a quiver and $F$ field. If $Q$ contains a path of length $|Q_0|$, then $FQ$ is infinite dimensional.

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

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

Proof: Let $a$ be an oriented loop in $Q$. Then $FQ$ is infinite dimensional by the above argument. Conversely, assume $Q$ 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 $|Q_0|-1$, there dimension of $FQ$ is bounded above by $\mathrm{dim}\,FQ\leq |Q_0|^2-|Q_0|<\infty$.

Lemma 13: Let $Q$ be a quiver and $F$ a field. Then the trivial edges $e_i$ form an orthogonal idempotent set.

Proof: This is immediate from the definitions: $e_ie_j=0$ if $i\neq j$ and $e_i^2=e_i$.

Corollary 14: The element $\sum_{i\in Q_0} e_i$ is the identity element in $FQ$.

Proof: It sufficed to show this on the generators of $FQ$. Let $a$ be a path in $Q$ with $s(a)=j$ and $t(a)=k$. Then $\left(\sum_{i\in Q_0} e_i\right)a=\sum_{i\in Q_0} e_ia=e_ja=a$. Similarily, $a\left(\sum_{i\in Q_0} e_i\right)=a$.

To be covered:

- General R-algebras