Constructing a field with exactly $81$ elements 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!!

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