 Field with four elements

The quotient by an irreducible polynomial is a field construction gives ${\mathbb{F}}_4={\mathbb{F}}_2[X]/(X^2+X+1)$. The quadratic has no root in ${\mathbb{F}}_2$. With $\alpha =[X]$, the four elements are $0,1,\alpha ,\alpha +1$, and ${\alpha }^2=\alpha +1$, $\alpha (\alpha +1)=1$, $(\alpha +1{)}^2=\alpha$. These relations determine the full multiplication table.