Associative algebra

Remarks

Equivalently, a $K$-algebra is a ring $A$ with a ring homomorphism $\eta \colon K \to A$ whose image lies in the center of $A$; the scalar product is then $ka = \eta(k)a$. Example: for a field $F$, the matrix ring $M_n(F)$ is an $F$-algebra with the entrywise scalar product. The check: $k(AB)$, $(kA)B$ and $A(kB)$ all have $(i,j)$ entry $k \sum_l a_{il} b_{lj}$, because $k$ commutes with the entries. Non-example: the quaternions $\mathbb{H}$ with left multiplication by elements of $\C \subset \mathbb{H}$ are not a $\C$-algebra. The condition $k(ab) = a(kb)$ fails for the scalar $i$ with $a = j$ and $b = 1$, since $i(j \cdot 1) = ij$ while $j(i \cdot 1) = ji = -ij \ne ij$. They are an algebra over $\R$, because real numbers commute with all quaternions. Every ring is a $\Z$-algebra in exactly one way. Some authors do not require a unit element in $A$; this note assumes one.