# Abelian groups from the definition of a field

1. Dec 4, 2007

### dodo

I see often repeated (in the Wikipedia page defining "Field", for one) that, from the field's axioms, it can be deduced that F,+ and F\{0},* are both commutative groups.

Yet, the closure property of * is only guaranteed on F, not necessarily on F\{0}. If I'm not mistaken, the finite field Z/6Z is a counter-example: 2*3=0 (mod 6), so {1,2,3,4,5} cannot be a group under multiplication. Is that correct?

2. Dec 4, 2007

### Hurkyl

Staff Emeritus
Z/6Z isn't a field...

3. Dec 4, 2007

### Coin

Hm, I think the trick here is that Z/6Z (assuming I understand your terminology... I think I would call this $$Z_6$$) is not a finite field. The only finite fields of this type are of order p, where p is a prime number. When p is prime the multiplicative group will always be closed under multiplication.

If you think about it, it would not make sense anyway for $$Z_6$$ to be a field. For example, if $$Z_6$$ is a field, then 2 would have to have a multiplicative inverse $$2^{-1}$$. But what do we get when we multiply $$2^{-1}$$, whatever it is, by 0? We get... 0, not 3 as we should. I think there are other violations of the field axioms as well. $$Z_6$$ is only an additive group.

4. Dec 4, 2007

### mathwonk

so as you observe, it must be a theorem that in a field, F-{0} is closed for *. i recall that as one of the first lemmas proved in abstract algebra. i believe it uses the fact that every non zero element has an inverse, and the definition of a unit for multiplication, plus the distributive property of multiplication. ohh, and the fact that 0 and 1 are different elements.

5. Dec 4, 2007

### dodo

Thanks for all your answers; that the order of a finite field has to have some restrictions (being a prime is one) makes now more sense.

I suppose the only think needed to show the closure of F\{0} under * is that, given two elements a,b of F, both non-zero, a*b cannot be zero. If a*b=0, since b is non-zero and therefore has an inverse b', then multiplying by b', a*b*b' = 0, and a=0, which contradicts the initial statement.

6. Dec 4, 2007

### morphism

I want to point out that the order of a finite field can be a power of a prime. However, Z/nZ is a field iff n is a prime. If p is a prime divisor of n but n/p isn't 1, then p*(n/p)=0 (mod n), i.e. Z/nZ admits a zero divisor. So Z/nZ can be a field only if n is a prime. The converse follows from the fact that nZ is a maximal ideal in Z when n is a prime (it's trivially a prime ideal and hence a maximal ideal because Z is a PID).

7. Dec 4, 2007

### dodo

It is a curious asymmetry; then there cannot be any isomorphism from a finite field F of order 9 to Z/9Z; if a is an element of F such that f(a)=3, a*a would be the 0 element of F. I'll try to look up examples of such finite field.

Last edited: Dec 4, 2007
8. Dec 4, 2007

### JasonRox

Well, 9 is not a prime therefore a finite field of order 9 does not exist.

9. Dec 5, 2007

### Hurkyl

Staff Emeritus
Except, of course, y'know, that one that does exist. :tongue:

$$GF(3^2) \cong \mathbf{Z}[x] / \langle 3, x^2 + 1 \rangle$$

Last edited: Dec 5, 2007
10. Dec 5, 2007

### dodo

11. Dec 5, 2007

### Hurkyl

Staff Emeritus
$\langle 3, x^2 + 1 \rangle$ is the ideal generated by 3 and x² + 1. So, the example you linked is a ring of reduced representatives for the one I posted.

(In fact, it turns out that all finite fields of order 9 are isomorphic)

12. Dec 5, 2007

### Coin

Yeah, when I said "The only finite fields of this type are of order p, where p is a prime number.", by "of this type" I meant the $$Z_n$$ fields, the ones obtained by applying a modulus to the integers. Sorry if I was unclear!

13. Dec 5, 2007

### JasonRox

Thanks for pointing that out.