Definition:Bilinear Mapping
![]() | This page has been identified as a candidate for refactoring of medium complexity. In particular: Equivalent definitions to be extracted and put into separate pages Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
Definition
Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct {A_1, +_1, \circ_1}_R, \struct {A_2, +_2, \circ_2}_R, \struct {A_3, +_3, \circ_3}_R$ be $R$-modules.
Let $\oplus: A_1 \times A_2 \to A_3$ be a binary operator with the property that: $\forall \tuple {a_1, a_2} \in A_1 \times A_2$:
- $a_1 \mapsto a_1 \oplus a_2$ 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, $\forall a, b \in R, \forall x, y \in A_2, z \in A_3$:
- $\paren {\paren {a \circ_1 x} +_1 \paren {y \circ_1 b} } \oplus z = \paren {a \circ_3 \paren {x \oplus z} } +_3 \paren {\paren {y \oplus z} \circ_3 b}$
and for all $z \in A_1, x,y \in A_2$:
- $z \oplus \paren {\paren {a \circ_2 x} +_2 \paren {y \circ_2 b} } = \paren {a \circ_3 \paren {z \oplus x} } +_3 \paren {\paren {z \oplus y} \circ_3 b}$
Equivalently, this can be expressed:
- $\paren {x +_1 y} \oplus z = \paren {x \oplus z} +_3 \paren {y \oplus z}$
- $z \oplus \paren {x +_2 y} = \paren {z \oplus x} +_3 \paren {z \oplus y}$
- $\paren {a \circ_1 x} \oplus z = a \circ_3 \paren {x \oplus z}$
- $z \oplus \paren {y \circ_2 b} = \paren {z \oplus y} \circ_3 b$
If $\struct {A, +, \circ}_R = A_1 = A_2 = A_3$, the notation simplifies considerably:
- $\paren {\paren {a \circ x} + \paren {b \circ y} } \oplus z = \paren {a \circ \paren {x \oplus z} } + \paren {b \circ \paren {y \oplus z} }$
- $z \oplus \paren {\paren {a \circ x} + \paren {y \circ b} } = \paren {a \circ \paren {z \oplus x} } + \paren {\paren {z \oplus y} \circ b}$
or equivalently, more easily digested:
- $\paren {x + y} \oplus z = \paren {x \oplus z} + \paren {y \oplus z}$
- $z \oplus \paren {x + y} = \paren {z \oplus x} + \paren {z \oplus y}$
- $\paren {a \circ x} \oplus z = a \circ \paren {x \oplus z}$
- $z \oplus \paren {y \circ b} = \paren {z \oplus y} \circ b$
Non-Commutative Ring
Let $R$ and $S$ be rings.
Let $M$ be a right $R$-module.
Let $N$ be a left $S$-module.
Let $T$ be an $\tuple {R, S}$-bimodule.
A bilinear mapping $f: M \times N \to T$ is a mapping which satisfies:
\(\text {(1)}: \quad\) | \(\ds \forall r \in R: \forall s \in S: \forall m \in M: \forall n \in N: \, \) | \(\ds \map f {r m, s n}\) | \(=\) | \(\ds r \cdot \map f {m, n} \cdot s\) | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \forall m_1, m_2 \in M : \forall n \in N: \, \) | \(\ds \map f {m_1 + m_2, n}\) | \(=\) | \(\ds \map f {m_1, n} + \map f {m_2, n}\) | ||||||||||
\(\text {(3)}: \quad\) | \(\ds \forall m \in M : \forall n_1, n_2 \in N: \, \) | \(\ds \map f {m, n_1 + n_2}\) | \(=\) | \(\ds \map f {m, n_1} + \map f {m, n_2}\) |