8

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?

user282177
  • 89
  • 2
    It is because if $n$ is not a power of a prime, then there does not exist a field with $n$ elements (you get in trouble with inverses). – Mankind Oct 20 '15 at 14:14
  • @HowDoIMath : The part of your comment that is not in parentheses appears to be only what the original poster had already read before posting here. The part in parentheses is the whole of the answer. ${}\qquad{}$ – Michael Hardy Oct 20 '15 at 18:09

3 Answers3

5

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 $0=p^{-1}\cdot0=p^{-1}pn'=n'$.

mathreadler
  • 25,824
CPM
  • 3,860
  • 17
  • 23
4

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.

Michael Hardy
  • 1
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)

Michael Hardy
  • 1
vudu vucu
  • 1,040