Associative algebras-some preliminary notes

Associative algebra:

Associative algebras are generalizations of field extensions and matrix algebras. For example, in field extension $E/F$,$E$ can be considered a $F$-algebra of dimension $n$. Also a $F$-vector space.

An associative algebra $\mathcal{A}$ is a ring, (with multiplication associative) with scalar multiplication and addition from a field $F$.

$K$-algebra means an associative algebra over field $K$.

In short, we want the $F$ action to be compatible with multiplication in $\mathcal{A}$. Say $f \in F$ and $a,b \in \mathcal{A}$ then
\begin{equation}
  (f.a)b=f.(ab)=a(f.b)
\end{equation}

We may consider $F$ as a subring under identification $f \rightarrow f.1_A$ where $1_A$ is multiplicative identity. Then, in compatibility condition noted above we can drop the dot in between $F$ elements and $\mathcal{A}$ elements
\begin{equation}
  fab = afb
\end{equation}
which is same as saying $fc=cf$ for some $c=ab$. Hence, this implies that $F \in Z(\mathcal{A})$ - that is in center of $\mathcal{A}$.



Examples:
A standard first example of a $K$-algebra is a ring of square matrices over a field $K$, with the usual matrix multiplication.

Let $F=Q$, the field of rationals. Consider the polynomial $X^2-2 \in Q[X]$. The splitting field $E=Q(\sqrt{2})$. Then $Q(\sqrt{2})=\{a+b\sqrt{2}|a,b\} \in Q$ is a vector space of $F$ over $E$ such that $dim_F(E)=2$.

Note, that an $n$-dimensional $F$-algebra $A$ can be realized as a subalgebra of $M_n(F)$ ($n\times n$ matrices over field $F$).

If $A,B$ are $F$-algebras, they can be added and multiplied via tensor operations. That is $A \otimes_F B$ and $A \oplus B$ are also associative algebras.

If $A$ is an algebra of dimension $2$, then $A \equiv F \oplus F$. This means $A$ is quadratic extension of $F$, or $A$ contains a nilpotent element.

To prove this, first we establish commutativity of $A$ using basis $\{1,\alpha\}$ over $F$.
 To see this simply expand $(x+y\alpha)(x'+y\alpha')$
 
 Quadratic extension requires that every non-zero element of $A$ should be invertible.

 Say $x+y\alpha$ be a non-zero element that is not invertible. This means $y \neq 0$ and can assert $\alpha$ is not invertible.

 $A$ can be represented via subalgebra of $M_2(F)$. So, we write $\alpha$ as
 \begin{align*}
   \alpha = [[a,b],[c,d]] \text{ a 2 by 2 matrix }
 \end{align*}
 Using above it is not too difficult to prove that $A \equiv F \oplus F$.

 Opposite algebras $A^{opp}$ is an algebra where multiplication is in reverse order. That is for $a,b \in A$, with $a.b$ as multiplication in $A$ and using $\times$ symbol for multiplication in $A^{opp}$, the condition is $b \times a = a.b$.