Field with four elements
= Field with four elements
{title2=$\mathbb F_4$}
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.