# Field with 27 Elements

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.

• In fact the elements in your field are all the polynomials of degree up to two in $\;\overline x:=\;$ the coset of $\;x\;$ in the quotient ring. – DonAntonio Feb 19 '16 at 15:50
• Sorry I meant to use $2x^3 + x + 2$. – Jmath99 Feb 19 '16 at 15:50
• The elements of the field are (equivalence classes of) polynomials of the shape $ax^2+bx+c$. As you wrote, there are three choices for each of $a,b,c$, total $27$. Why the $2x^3$, it would be more pleasant to multiply your cubic by $2$. – André Nicolas Feb 19 '16 at 15:52
• By the way, the elements are of the form $ax^2+bx+c$, not $ax^3+bx+c$ – ASKASK Feb 19 '16 at 15:52
• Another way to find the number of elements is to view the residue field ${\Bbb Z}_3 [X] / \langle 2x^3 + x +2 \rangle$ as a vector space over ${\Bbb Z}_3$ with basis $1,x,x^2$, so it has $27$ elements. – user60589 Feb 19 '16 at 15:57

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.

• Strictly speaking, one should also argue that such an expression is unique. – Manny Reyes Feb 19 '16 at 17:50

It must be said that an alternative way to build a finite field with $p^n$ elements (they all have this number of 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), it suffices 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.

See the elementary article of W. P. Wardlaw in Mathematics Magazine, Oct. 1994, that can be found on the net (Wardlaw47052.pdf).