Field with $125$ elements 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 $\Bbb F_5[x]$?

Belatedly, there is a way to "be lucky" here, and not so computational, in an informative way. Namely, of all the polynomials known to humans, the best understood are cyclotomic ones. If a cyclotomic polynomial or a relative can resolve an issue, that's good fortune.
After a moment's fooling around, we observe that the smallest power of $5$ such that $5^n-1$ is divisible by $7$ is $5^6$. That is, $5$ happens to be a "primitive root" mod $7$. Thus, a seventh root of unity $\zeta$ is of degree $6$ over $\mathbb F_5$. That is, the seventh cyclotomic polynomial $x^6+x^5+...+x+1$ is irreducible over $\mathbb F_5$.
The irreducible of $\xi=\zeta+\zeta^{-1}$ over $\mathbb F_5$ is obtained by a standard algebraic manipulation:
$$
0\;=\; 
\zeta^{-3}\Big(\zeta^6+\ldots+\zeta+1\Big)
\;=\;
\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}
$$
$$
\;=\; (\xi^3 - 3\xi) + (\xi^2-2) + 1
\;=\; \xi^3 + \xi^2 - 3\xi - 1
$$
We can use $-3=2$ and $-1=4$ if we like.
So, then, basic Galois theory assures us that $x^3+x^2-3x-1$ is irreducible in $\mathbb F_5[x]$.
