Algebra over a field

Remarks

The three conditions say that the product is $K$-bilinear. So an algebra over $K$ is a nonassociative algebra over the ring $K$. The algebra is associative if $(xy)z = x(yz)$ for all $x, y, z$, commutative if $xy = yx$ for all $x, y$, and unital if it has an element $1$ with $1x = x1 = x$ for all $x$. Example: $\C$ is an algebra over $\R$; complex multiplication distributes over addition and $(ax)(by) = ab\,xy$ for real $a, b$. Non-example: on $\R^2$, the product $(x_1, x_2) * (y_1, y_2) = (x_1 y_1, 1)$ fails clause 3, since $(0 \cdot x) * (0 \cdot y) = (0, 1) \neq (0, 0)$.