Field with nine elements
= Field with nine elements
{title2=$\mathbb F_9$}
The <finite field> $\mathbb F_9$ can be represented as $\mathbb F_3[X]/(X^2+1)$, since the quadratic has no root modulo three. Its elements are $a+b\beta$ for $a,b\in\mathbb F_3$, with $\beta^2=-1$. Multiplication reduces to $(a+b\beta)(c+d\beta)=(ac-bd)+(ad+bc)\beta$ modulo three.