Field with 27 Elements

Question

I am trying to construct a field with 27 elements. So far I have found an irreducible polynomial of degree 3, $2x^3 + x + 2$ in $\mathbb{Z}_3$ and thus $\langle 2x^3 + x + 2 \rangle$ is maximal in $\mathbb{Z}_3$. Now all that remains is for me to prove the field $\mathbb{Z}_3/ \langle 2x^3 + x + 2 \rangle$ has 27 elements. I know the elements in this field look like $ax^2 + bx + c$ + $f(x)$ with 3 choices for each constant term in the left coset. All that remains is for me to provide justification that The example I provided indeed has 27 elements. Any help would be appreciated on how to do so.

Answer 1

If every element in the field is of the form $ax^2+bx+c + <f(x)>$ and there are three choices for $a$, three choices for $b$, and three choices for $c$, and different choices give rise to different elements, then by some very elementary combinatorics there are $3 \times 3 \times 3=27$ possible choices.

Answer 2

An alternative way to build a finite field with $p^n$ elements (with $p$ prime) is by using certain $n \times n$ matrices with coefficients in $\mathbb{Z/pZ}$. In fact, due to Wedderburn theorem (the multiplicative group of a finite field is cyclic), one has just to find a matrix $G$ generating a cyclic group with $p^n-1$ elements (here $p^n-1=26$: all powers $G^k$ from $k=0$ to $k=25$ are different).

One of them, among many others, is obtained as a companion matrix

$$G=\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix} \ \ \ \ \ (1)$$

There would be much more to say, in particular regarding the isomorphism with the polynomial construction. This isomorphism uses the irreducibility of the characteristic polynomial of matrix $G$ ; in the case of matrix $G$ given by (1), its characteristic polynomial is a multiple of $2x^3+x+2$, the example you have given.

Remarks : 1) A way to produce such a generating matrix is by considering the matrix associated with multiplication

2) See the elementary article of W. P. Wardlaw in Mathematics Magazine, Oct. 1994