## Exhibit some finite fields

Let $g(x) = x^2+x-1$ and $h(x) = x^3-x+1$. Construct finite fields of order 4, 8, 9, and 27 elements. For the fields with 4 and 9 elements, give the multiplication tables and show that the nonzero elements form a cyclic group.

Note that to construct fields of the given sizes, it suffices to show that $g(x)$ and $h(x)$ are each irreducible over the fields $\mathbb{F}_2$ and $\mathbb{F}_3$. (More precisely, that these polynomials are irreducible when we inject the coefficients into the respective fields.)

It is easy to see that neither $g$ nor $h$ has a root in either $\mathbb{F}_2$ or in $\mathbb{F}_3$, so these are indeed irreducible. Then $\mathbb{F}_2[x]/(g(x))$, $\mathbb{F}_2[x]/(h(x))$, $\mathbb{F}_3[x]/(g(x))$, and $\mathbb{F}_3[x]/(h(x))$ are fields of order 4, 8, 9, and 27 by Proposition 11 in D&F, and specifically are isomorphic as fields to the extension of $\mathbb{F}_2$ or $\mathbb{F}_3$ by adjoining a root of $g$ or $h$.

The multiplication table of $\mathbb{F}_2(\alpha) \cong \mathbb{F}_2[x]/(g(x))$ is as follows.

 $\begin{array}{c|cccc} & 0 & 1 & \alpha & \alpha+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & \alpha+1 \\ \alpha & 0 & \alpha & \alpha+1 & 1 \\ \alpha+1 & 0 & \alpha+1 & 1 & \alpha \end{array}$

Evidently, $(\mathbb{F}_2(\alpha))^\times$ is generated by $\alpha$.

The multiplication table of $\mathbb{F}_3(\alpha) \cong \mathbb{F}_3[x]/(g(x))$ is as follows.

 $\begin{array}{c|ccccccccc} & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ 2 & 0 & 2 & 1 & 2\alpha & 2\alpha+2 & 2\alpha+1 & \alpha & \alpha+2 & \alpha+1 \\ \alpha & 0 & \alpha & 2\alpha & 2\alpha+1 & 1 & \alpha+1 & \alpha+2 & 2\alpha+2 & 2 \\ \alpha+1 & 0 & \alpha+1 & 2\alpha+2 & 1 & \alpha+2 & 2\alpha & 2 & \alpha & 2\alpha+1 \\ \alpha+2 & 0 & \alpha+2 & 2\alpha+1 & \alpha+1 & 2\alpha & 2 & 2\alpha+2 & 1 & \alpha \\ 2\alpha & 0 & 2\alpha & \alpha & \alpha+2 & 2 & 2\alpha+2 & 2\alpha+1 & \alpha+1 & 1 \\ 2\alpha+1 & 0 & 2\alpha+1 & \alpha+2 & 2\alpha+2 & \alpha & 1 & \alpha+1 & 2 & 2\alpha \\ 2\alpha+2 & 0 & 2\alpha+2 & \alpha+1 & 2 & 2\alpha+1 & \alpha & 1 & 2\alpha & \alpha+2 \end{array}$

Evidently, $(\mathbb{F}_3(\alpha))^\times$ is generated by $\alpha$.