 Field with nine elements

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.