… as the binary operation
\[\begin{gather*} \forall a, b \in F, \\[1ex] ab \in F. \end{gather*} \]
where | \(a, b\) | are factors, |
\(ab\) | is the product, and | |
\(F\) | is the field with \(a\) and \(b\). |
… by any of the
\[\begin{equation*} ab = a \cdot b = a \times b. \end{equation*} \]
… is \(1\), so
\[\begin{gather*} \forall a \in F, \\[1ex] 1a = a. \end{gather*} \]
More generally,
\[\begin{gather*} \forall a \in F, \> \exists! b \in F : \\[1ex] a b = a, \end{gather*} \]
where the identity is
\[\begin{equation*} b = 1. \end{equation*} \]
… is \(a^{-1}\), so
\[\begin{gather*} \forall a \neq 0 \in F, \\[1ex] a a^{-1} = 1. \end{gather*} \]
More generally,
\[\begin{gather*} \forall a \neq 0 \in F, \> \exists! b \in F : \\[1ex] a b = 1, \end{gather*} \]
where the inverse is \(b = 1 / a = a^{-1}.\)
… holds, so
\[\begin{gather*} \forall a, b \in F, \\[1ex] ab = ba. \end{gather*} \]
… holds, so
\[\begin{gather*} \forall a, b, c \in F, \\[1ex] (ab)c = a(bc). \end{gather*} \]
… holds over addition, so
\[\begin{gather*} a, b, c \in F, \\[1ex] c(a + b) = ca + cb \end{gather*} \]
which is the one and only
link between addition and multiplication
in the field.