Constructing a field with exactly $81$ elements

Question

I was thinking $\frac{\mathbb{Z_3}[x]}{(x^3+x+1)} \times \frac{\mathbb{Z_3}[x]}{(x^3+x+1)}$.

$(x^3+x+1)$ is irreducible in $\mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!

How would I do this if working over $\mathbb{Z_9}$ and $\mathbb{Z_2}$? in $\mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!

Answer 1

No you can't do $F \times F$ because $(a,0) \times (0,b) = (0,0)$ would break the field axioms.

You need to give an irreducible polynomial $P$ of degree $4$ over $\mathbb{Z}_3$ so you can give representatives of $\mathbb{Z}_3[x]/(P)$ by polynomials of degree $\leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.

Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test

Answer 2

A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $\mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.

Answer 3

Let's see. What are the irreducible quadratics: $x^2+x+2\,,x^2+2x+2\,,x^2+1$.

I count $3$ (thanks @Lubin).

So, there are ${3\choose 2}+3=6$ combinations to check.

These are the $6$ quartics I get: $x^4+1\,,x^4+x^3+x+1\,,x^4+2x^3+2x+2 \,,x^4+2x^3+x^2+x+1\,,x^4+x^3+x^2+2x+1\,,x^4+2x^2+1$.

But there are $2\cdot 3^4=162$ quartics to choose from... $81$ of them monic.

So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.

So, how about $x^4+2x^3+2$?

Finally, $\mathbb F_{3^4}\cong\frac{\mathbb Z_3[X]}{x^4+2x^3+2}$.

Answer 4

If you don't want to mess around with finding an irreducible quartic, you could construct $\mathbb{F}_{81}$ as the quotient of $\mathbb{F}_9[x]$ with a quadratic irreducible in $\mathbb{F}_9$. $\mathbb{F}_9$ can be constructed as the quotient of $\mathbb{F}_3[y]$ with a quadratic irreducible in $\mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.

Expressing $\mathbb{F}_9\cong\mathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $\mathbb{F}_9$, so $\mathbb{F}_{81}\cong\mathbb{F}_9[x]/(x^2+y+1)$.