Answer to Question #179099 in Abstract Algebra for Anand

Question

Construct a field with 125 elements.

Answer 1

Because $125=5^3$ we need to find an irreducible polynomial of the third degree in the ring $\mathbb{Z}_5[X]$ and perform factorization (because if we perform factorization on irreducible polynomial, we get the field. In this field, the "polynomials" will be at most of the second degree and each coefficient will be from $\mathbb{Z}_5$. From combinatorial considerations it is clear that the power of the field will be 125)

A suitable polynomial $x^3+x^2+x+4$

So: $\mathbb{Z}_5[X]/<X^3+X^2+X+4>\ \simeq \mathbb{F}_{125}$