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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.

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    $\begingroup$ Why "..., but clearly $1\cdot x=x\cdot 1=x\in A$"? Is it obvious? $\endgroup$ – Song Jan 9 at 12:13
  • $\begingroup$ If $A$ is not all of $R$, then there would exist some element $a_k \in R$ such that $ca_i = a_k=ca_j$ $\endgroup$ – JavaMan Jan 9 at 12:27
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    $\begingroup$ 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) $\endgroup$ – Ovi Jan 9 at 12:28
  • $\begingroup$ That's what I was worried about, what can I do to show that R is a subset of A? $\endgroup$ – mathsssislife Jan 9 at 12:30
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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.

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  • $\begingroup$ Is there a way to do it without the need to show that the map is injective? $\endgroup$ – mathsssislife Jan 9 at 15:55
  • $\begingroup$ @mathsssislife not that I know $\endgroup$ – Anguepa Jan 10 at 1:15

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