1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DonAntonio Feb 19 '16 at 15:50
  • $\begingroup$ Sorry I meant to use $2x^3 + x + 2$. $\endgroup$ – Jmath99 Feb 19 '16 at 15:50
  • $\begingroup$ 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 $2$. $\endgroup$ – André Nicolas Feb 19 '16 at 15:52
  • $\begingroup$ By the way, the elements are of the form $ax^2+bx+c$, not $ax^3+bx+c$ $\endgroup$ – ASKASK Feb 19 '16 at 15:52
  • $\begingroup$ 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. $\endgroup$ – user60589 Feb 19 '16 at 15:57
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Strictly speaking, one should also argue that such an expression is unique. $\endgroup$ – Manny Reyes Feb 19 '16 at 17:50
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.