TITLE: Construct a finite field of order 27
QUESTION [5 upvotes]: So some of my thoughts for constructing a finite field of order 27 are making me think of a field with $p^n$ elements, where $p = 3$ and $n = 3$ such that we want a cubic polynomial in $\mathbb{F}_3[X]$ that does not factor. 
Could this be thought of as looking for a cubic polynomial in $\mathbb{F}_3[X]$ with no roots in $\mathbb{F}_3$? Could this polynomial work: $x^3 + 2x^2 + 1$ ?

REPLY [8 votes]: Yes, it does work: it is irreducible because it has no roots in $\mathbb{Z}_3$ (and $\mathbb{Z}_3$ is a field). Thus, the quotient ring $\mathbb{Z}_3[x]/(x^3 + 2x^2 +1)$ is a field which has $3\cdot 3\cdot 3$ elements.