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.

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