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.

• Why "..., but clearly $1\cdot x=x\cdot 1=x\in A$"? Is it obvious? – Song Jan 9 at 12:13
• If $A$ is not all of $R$, then there would exist some element $a_k \in R$ such that $ca_i = a_k=ca_j$ – JavaMan Jan 9 at 12:27
• I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial) – Ovi Jan 9 at 12:28
• That's what I was worried about, what can I do to show that R is a subset of A? – mathsssislife Jan 9 at 12:30

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.