Finite Fields. Why does it involve prime numbers only?

Question

I'm just getting my head around the finite fields, so called Galois Fields. Why are they based on prime numbers only? any concept I'm missing?

Answer 1

All fields must have characteristic $0$ (like $\mathbb{Q}$ or $\mathbb{R}$) or else characteristic $p$ a prime. If the characteristic were $n$, a composite, say $n=pn'$, then $0=pn'$. This shows $p$ is a zero-divisor and therefore not a unit.

If $p$ were a unit, you would have $0=p^{-1}\cdot0=p^{-1}pn'=n'$.

Answer 2

Assume $\operatorname{char} F=n=ab>0$ with $a,b>1$. Then $0=n=(1+\cdots+1)=(1+\cdots+1)(1+\cdots+1)=ab$.

Note that $a,b \in F\setminus\{0\}$. Does it happen $ab=0$ in field?

see: characteristic (algebra)

Answer 3

Let's try it with a composite number, $12$. We have $3\cdot 4 = 0$, so $$ 0 = 3^{-1}\cdot 0 = 3^{-1}\cdot(3\cdot4) = (3^{-1}\cdot3)\cdot 4 = 1\cdot4 = 4. $$

That's why it does not work with composite numbers.

Why it does work with prime numbers is a more substantial thing: How do you prove everything except $0$ has a multiplicative inverse in that case? That involves the quotients in Euclid's algorithm.