Let \(F\) be a set which contains exactly two elements, \(0\) and \(1\). Define an addition and multiplication by the tables \[ \begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \end{array} \qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \end{array} .\] Verify that the set \(F\), together with these two operations, is a field.
Solution
1. Addition is commutative and associative, with identity \(0\) and each element its own inverse.
the table is symmetric; \(0 + a = a\); \(1 + 1 = 0\) so \(1\) is its own inverse.
2. Multiplication is commutative and associative, with identity \(1\).
the table is symmetric; \(1 \cdot a = a\); associativity is checked directly (the only nontrivial case is \(1 \cdot 1 \cdot 1 = 1\)).
3. Multiplication distributes over addition.
\(a(b + c) = ab + ac\) holds for all \(a, b, c \in \{0, 1\}\) by direct check (the only case needing care is \(1 \cdot (1 + 1) = 1 \cdot 0 = 0 = 1 + 1 = 1 \cdot 1 + 1 \cdot 1\)).
4. Every nonzero element has a multiplicative inverse.
the only nonzero element is \(1\), and \(1 \cdot 1 = 1\).
5. Hence \(F\) is a field.
<1>1–<1>4 verify the field axioms; this is the field \(\FF_2\) with two elements.
6. Q.E.D.
5.