Talk:Finite Integral Domain is Galois Field/Proof 2

I read the proof and it looks ok, so I removed the read proof tag.

Which are the missing links?

I have been told that no matter how advanced a proof is that it is better to have the links present in an article than not be so. This differentiates ProofWiki's proofs from a normal proof on the net which assumes you have a particular set of knowledge already. --Jshflynn (talk) 14:29, 19 September 2012 (UTC)

Actually this proof is not okay, it's got a hole in the middle.

It can be improved by:

a) Call the integral domain $\left({D, +, \circ}\right)$ or whatever symbols you want to use for the operations. It might make it easier to follow the proof.

b) Treat the group $\left({R^*, \circ}\right)$ as a separate entity, where $R^* = R \setminus \left\{{0}\right\}$, that is, $R$ without $0$.

c) Establish that $\left({R^*, \circ}\right)$ is in fact a group, at which stage the proof is finished:

i) As $R$ is an integral domain, it has no zero divisors. Therefore $R^*$ is closed under $\circ$. (This is the hole I was talking about.)

ii) By definition, $\left({R^*, \circ}\right)$ is a semigroup whose identity is $1$.

iii) The rest of the proof hangs together but it's a little messy: cancellation law proves injectivity and (as in proof 1) pigeonhole principle proves surjectivity.

iv} So the left regular representation (if you want to go down that route) is a bijection and so a permutation and hence the result from whatever group theory result proves whatever it was we were trying to prove in the first place.

v) Finally you need to mention that commutativity of $R$ means that $\left({R^*, \circ}\right)$ is abelian. Commutativity is mentioned in this proof, but not with this particular relevance.

Steady as she goes Mr. Sulu. --prime mover (talk) 19:26, 19 September 2012 (UTC)