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)