# Construct a finite field of order 27

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: $x^3 + 2x^2 + 1$ ?

• why is your choice $x^3+2x^2+1$?? Was this a hint or you somehow felt this would work? – user87543 Dec 9 '13 at 15:38
• This was a hint – user110655 Dec 9 '13 at 15:39
• What can you say about $R/M$ where $R$ is a commutative ring with unity and $M$ is a maximal ideal. In a PID, what can you say about the relationship between irreducible elements, prime ideals and maximal ideals? – LASV Dec 9 '13 at 15:42

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.