Taken from Hopf Algebras by Abe. I am not doing this over a commutative ring as he is doing but just over the field of complex numbers, $\mathbb{C}$ — therefore some of my constructions will be simplified. Some of them might even be wrong.

In this section the notion of algebras will be introduced; and we will present some examples of such algebras. An algebra over $\mathbb{C}$ is defined by giving a structure map.

### Algebras

Suppose $A=\{e_\lambda,0\}_{\lambda\in\Lambda}$ is a ring and we are given a ring morphism $\eta_A:\mathbb{C}\rightarrow A$. Then $A$ can be viewed as a complex vector space. If we have $(ax)y=x(ay)=a(xy)$ for all $a\in\mathbb{C}$ and $x,\,y\in A$

then $A$ is said to be an algebra. Now the map defined by $f(x,y)=xy$ turns out to be bilinear. Thus we obtain a morphism $\nabla_A:A\otimes A\rightarrow A$.

From this fact, we see that an algebra $A$ can be defined in the following manner. For a complex vector space $A$ and morphisms $\eta_A:\mathbb{C}\rightarrow A$ $\nabla_A:A\otimes A\rightarrow A$, we have the following: $\nabla_A\circ(\nabla_A\otimes I_A)=\nabla_A\circ(I_A\otimes\nabla_A)$

(the associative law) $\nabla_A\circ(\eta_A\otimes I_A)=\nabla_A\circ (I_A\otimes \eta_A)=I_A$

(the unitary property)

Here, $\nabla_A$ is said to be the multiplicative map of $A$ $\eta_A$ the unit map of $A$, and together we call $\nabla_A$ $\eta_A$ the structure maps of the algebra $A$.

Given algebras $A$ and $B$, if a map $\varphi:A\rightarrow B$ is a homomorphism as well as a linear map, then we call $\varphi$ an algebra morphism. For algebras $A$ $B$, if we let $\nabla_A,\,\nabla_B,\,\eta_A,\,\eta_B$ be their respective structure maps, then the linear map is an algebra morphism if and only if we have $\nabla_B\circ(\varphi\otimes\varphi)=\varphi\circ \nabla_A$

and $\eta_B=\varphi\circ \eta_A$.

Given algebras $A$ and $B$, the vector space $A\otimes B$ becomes an algebra where the structure maps are given by $\nabla_{A\otimes B}=(\nabla_A\otimes\nabla_B)(1_A\otimes \tau\otimes1_B)$, $\eta_{A\otimes B}=\eta_A\otimes \eta_B$.

The map $\tau$ above denotes the flip map: a vector space isomorphism $A\otimes B\rightarrow B\otimes A$ defined by $x\otimes y\mapsto y\otimes x$. The algebra $A\otimes B$ is called the tensor product of $A$ and $B$. The multiplication in $A\otimes B$ is given by $(a_1\otimes b_1)(a_2\otimes b_2)=a_1a_2\otimes b_1b_2$, for $a_i\in A$ $b_i\in B$,

and the unit element is given by $1_A\otimes 1_B$.

### Example 1.3

Given a set of indeterminates $\{x_\lambda\}_{\lambda\in\Lambda}$, the set of all polynomials in $x_\lambda$ ( $\lambda\in\Lambda$) with coefficients in $\mathbb{C}$ is written $\mathbb{C}[x_\lambda]_{\lambda\in\Lambda}$ is a commutative ring via the addition and multiplication defined naturally. Moreover, the canonical embedding $\mathbb{C}\rightarrow\mathbb{C}[x_\lambda]_{\lambda\in\Lambda}$ makes $\mathbb{C}[x_\lambda]_{\lambda\in\Lambda}$ an algebra (i.e. the embedding is the unit map).

### Example 1.4

The set $M_n(\mathbb{C})$ of all $n\times n$ square matrices with complex coefficients is a ring with respect to addition and multiplication of matrices and is a vector space with respect to scalar multiplication. Moreover, $M_n(\mathbb{C})$ becomes an algebra when we define $\eta_{M_n(\mathbb{C})}:a\mapsto aI_{n\times n}$.

### Example 1.5

Let $A=F(S)$ the set of all functions from a set $S$ to $\mathbb{C}$. For $f\,g\in A$, we define the operations $(f+g)(x)=f(x)+g(x)$ $fg(x)=f(x)g(x)$ $x\in S$,

which makes $A$ a ring. Defining the action $(kf)(x)=k\cdot f(x)$ $k\in\mathbb{C}$ $x\in S$, $A$ becomes a vector space. Notice that $A$ becomes an algebra when we define a vector space morphism $\eta_A(k)(x)=k$ $x\in A$.

### Example 1.6

Let $G$ be a group and denote the the free complex vector space (I’m restricting to complex vector spaces rather than modules.) generated by $G$ as $\mathbb{C}G$. Defining the multiplication on $\mathbb{C}g$ by $\left(\sum_{x\in G}a_xx\right)\left(\sum_{y\in G}b_yy\right)=\sum_{z\in G}\left(\sum_{xy=z}a_xb_y\right)z$, $\mathbb{C}G$ becomes a ring. The map $\eta_{\mathbb{C}G}:\mathbb{C}\rightarrow\mathbb{C}G$ given by $\eta_{\mathbb{C}G}(k)=ke_G$, where $e_G$ is the identity element of $G$, makes $\mathbb{C}G$ into an algebra. The algebra $\mathbb{C}G$ is said to be the group algebra of $G$.