I want to construct a field with $125$ elements. My idea is to consider the polynomial ring $\Bbb F_5[x]$. It is enough to find an irreducible polynomial $f\in \Bbb F_5[x]$ of degree $3$ because then $\Bbb F_5[x]/(f)$ is a field with exactly $5^3=125$ elements.
How do I find an irreducible polynomial of degree $3$ in $\Bbb F_5[x]$?
Best Answer
If a cubic polynomial of $\mathbb{F}_5[x]$ is reducible, then it splits into a linear factor and a quadratic factor or into the product of three linear factors. Linear factors are very easy to test for, as $(x-a)$ is a factor of $f$ if and only if $f(a) = 0$.
So you might choose a random degree $3$ polynomial and test for the five possible roots.
For instance, I'm tempted to try $f(x) = x^3 + 2x + 1$. Then $$\begin{align} &f(0) = 1, \\ &f(1) = 4, \\ &f(2) = 8 + 4 + 1 \equiv 3, \\ &f(3) = 27 + 6 + 1 \equiv 4, \\ &f(4) = 64 + 8 + 1 \equiv 3. \end{align}$$
As none of these are zero, $f(x)$ is an irreducible cubic.
Note that testing for roots explicitly doesn't work for higher degree polynomials. It is possible that a reducible quartic factors into the product of two quadratics. So though there are no roots, it might still be reducible. However since we know that a reducible cubic has a linear factor, we can test just for the linear factor.