Lemma: Let R be a finite integral domain. Assume the elements of R are {$a_1$,$a_2$,,,,$a_n$} for some n$\in$ $\mathbb{N}$. Then for c$\in$R, A={c$a_1$,c$a_2$,,,,c$a_n$}=R.
My proof: Since R is a finite integral domain. By definition, R is a finite commutative ring with the property that $\forall$ a,b,c $\in$ R with c$\neq0$ if $ca=cb$ then $a=b$. Let R have elements {$a_1$,$a_2$,,,,$a_n$}. Consider the set A={c$a_1$,c$a_2$,,,,c$a_n$}, for some c$\in$R, then each element is distinct. However; since by the definition of a ring, the ring is closed under multiplication, it follows that A$\subseteq$R. Now, we must show that R$\subseteq$A. Since $1.x=x.1=x$, it follows that x$\in$A; therefore, R$\subseteq$A.
I am trying to avoid the theorem that every finite integral domain is a field, partly because i'm using the above lemma to prove that every finite integral domain is a field.
Is my proof correct? What can I do to make it better? I'm just worried about the 1.x =x.1 =x part because I assumed that it holds for an arbitrary c and now i'm assuming it holds for c=1.
Of course this only works if $c\neq 0$. You must show that the map $x \mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2\mathbb{Z}$ are a proper subset of the integers $\mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.