How to construct a field of order 8 and 9

Question

Construct two fields, one of order 8 and one of order 9.

I only know that the field has 8 elements in the former case.

Answer 1

Note that $8=2^3$ and $9=3^2$. The first thing that needs to be done is to find an irreducible cubic polynomial in GF(2) for the order-8 case, and an irreducible quadratic polynomial in GF(3) for the order-9 case; $x^3-x-1$ and $x^2+1$ will do.

Define $\alpha$ and $\beta$ with $\alpha^3=\alpha+1$ and $\beta^2=-1$. Then the elements of the order-8 field are $p+q\alpha+r\alpha^2$ where $p,q,r\in\{0,1\}$. The elements of the order-9 field are $s+t\beta$ with $s,t\in\{0,1,2\}$.

Answer 2

Here is why Parcly's answer works: Let $F$ be a field (like $\mathbb R$, $\mathbb C$, $\mathbb F_p$,...). Let $x$ be a variable, so that the powers $x^k$ are independent of each other. Then you get a ring $F[x]$ of all polynomials with coefficients over the field $F$. Note that $F[x]$ is a PID (and therefore a UFD) since it is a Euclidean domain via polynomial division with remainder. The upshot here is that irreducible elements in $F[x]$ generate maximal ideals.

Since the polynomials $x^3 - x - 1$ and $x^2 + 1$ are irreducible over $\mathbb F_3$ (for they don't have roots), they are irreducible elements in $\mathbb F_3[x]$, whence the ideals they generate are maximal. If $f$ is one of these polynomials (or something similar), then $\mathbb F_3[x]/(f)$ is a field by the ideal correspondence theorem.