\begin{definition}[Definition:Bilinear Mapping]
Let $\left({R, +_R, \times_R}\right)$ be a commutative ring.
Let $\left({A_1, +_1, \circ_1}\right)_R, \left({A_2, +_2, \circ_2}\right)_R, \left({A_3, +_3, \circ_3}\right)_R$ be $R$-modules.
Let $\oplus: A_1 \times A_2 \to A_3$ be a binary operator with the property that: $\forall \left({a_1, a_2}\right) \in A_1 \times A_2$:
: $a_1 \mapsto a_1 \oplus a_2$ is a linear transformation from $A_1$ to $A_3$
: $a_2 \mapsto a_1 \oplus a_2$ is a linear transformation from $A_2$ to $A_3$
Then $\oplus$ is a '''bilinear mapping'''.
That is, $\forall a, b \in R, \forall x, y \in A_2, z \in A_3$:
: $\left({\left({a \circ_1 x}\right) +_1 \left({y \circ_1 b}\right)}\right) \oplus z = \left({a \circ_3 \left({x \oplus z}\right)}\right) +_3 \left({\left({y \oplus z}\right) \circ_3 b}\right)$
and for all $z \in A_1, x,y \in A_2$:
: $z \oplus \left({\left({a \circ_2 x}\right) +_2 \left({y \circ_2 b}\right)}\right) = \left({a \circ_3 \left({z \oplus x}\right)}\right) +_3 \left({\left({z \oplus y}\right) \circ_3 b}\right)$
Equivalently, this can be expressed:
: $\left({x +_1 y}\right) \oplus z = \left({x \oplus z}\right) +_3 \left({y \oplus z}\right)$
: $z \oplus \left({x +_2 y}\right) = \left({z \oplus x}\right) +_3 \left({z \oplus y}\right)$
: $\left({a \circ_1 x}\right) \oplus z = a \circ_3 \left({x \oplus z}\right)$
: $z \oplus \left({y \circ_2 b}\right) = \left({z \oplus y}\right) \circ_3 b$
If $\left({A, +, \circ}\right)_R = A_1 = A_2 = A_3$, the notation simplifies considerably:
: $\left({\left({a \circ x}\right) + \left({b \circ y}\right)}\right) \oplus z = \left({a \circ \left({x \oplus z}\right)}\right) + \left({b \circ \left({y \oplus z}\right)}\right)$
: $z \oplus \left({\left({a \circ x}\right) + \left({y \circ b}\right)}\right) = \left({a \circ \left({z \oplus x}\right)}\right) + \left({\left({z \oplus y}\right) \circ b}\right)$
or equivalently, more easily digested:
: $\left({x + y}\right) \oplus z = \left({x \oplus z}\right) + \left({y \oplus z}\right)$
: $z \oplus \left({x + y}\right) = \left({z \oplus x}\right) + \left({z \oplus y}\right)$
: $\left({a \circ x}\right) \oplus z = a \circ \left({x \oplus z}\right)$
: $z \oplus \left({y \circ b}\right) = \left({z \oplus y}\right) \circ b$
\end{definition}