# Construction of non-prime finite fields

I am new to Galois field theory and I am struggling with some definitions. To construct any non-prime finite field $GF(p^n)$ with p prime and $n \in \mathbb{N}$, one has to find an irreducible polynomial $g(x)$ in $GF(p)$ and eventually calculate $GF(p^n) = G(p)[x] / g(x)$.

Assuming I want to construct $GF(9) = GF(3^2)$. Why do I have to do the stuff above? Doesn't $GF(3^2)$ simply contains elements ranging from 0 to 8? What is the upper construction rule about?

• No, $GF(9)$ is very different from $\Bbb Z/9\Bbb Z$. The latter has zero divisors, for example. – Gregory Grant May 31 '15 at 10:27
• Take $\Bbb Z_3[x]/(x^2 + 1)$, for instance. It can alternately be described using complex numbers: $$\{a + bi \mid a, b \in \Bbb Z_3\}$$with addition and multiplication defined as with the usual complex numbers, but with reducing modulo $3$ afterward. – Arthur May 31 '15 at 10:30
• But don't you have to use an irreducible polynomial of degree 3? – null May 31 '15 at 10:33
• @null If you quotient by an irreducible polynomial of degree 3 then the resulting field would have $3^3$ elements. More generally, if you quotient by a degree $d$ polynomial, the resulting field would have $3^d$ elements. – man and laptop May 31 '15 at 10:35

The only finite fields that equal the rings $\Bbb Z/n\Bbb Z$ are the ones where $n$ is a prime. Because if $n=ab$ then $\overline{a}\cdot\overline{b}=\overline{0}$ in $\Bbb Z/n\Bbb Z$ and we know that can't happen in a field. On the other hand if you adjoin a root of an irreducible polynomial of degree $2$ to $\Bbb Z/3\Bbb Z$ then you get a degree two extension of $\Bbb Z/3\Bbb Z$ which has nine elements, but is not equal to $\Bbb Z/9\Bbb Z$.