Prove R is a Field

Question

Let R be a integral domain with a finite number of elements. Prove that R is a field.

Let a ∈ R \ {0}, and consider the set aR = {ar : r ∈ R}.

Guessing i will have to show that |aR| = R, and deduce that there exists r ∈ R such that ar = 1 but don't know what to do?

score 2 · Accepted Answer · answered Oct 18 '16 at 13:18

2

HINT: If $aR\ne R$, there must be distinct $r,s\in R$ such that $ar=as$. (Why?) But then $a(r-s)=\ldots\;?$

answered Oct 18 '16 at 13:18

Brian M. Scott

616,228

i see what you're saying @brian but how will this help? – Sam Harper Oct 18 '16 at 13:22
@SamHarper If $aR = R$ for all $a \in R$, then $ab = 1$ for some $b \in R$. – J126 Oct 18 '16 at 13:23
@Sam: Did Joe’s comment take care of your question, or should I say more? – Brian M. Scott Oct 18 '16 at 13:24
sorry I'm not sure i understand @BrianM.Scott – Sam Harper Oct 18 '16 at 13:30
@Sam: $a(r-s)$ would be $0_R$, but $R$ has no zero-divisors, so this is impossible. Therefore the map $r\mapsto ar$ must be an injection, and since $R$ is finite, it must be a bijection. Thus, $aR=R$. But that means that there is some $b\in R$ such that $ab=1_R$. – Brian M. Scott Oct 18 '16 at 13:33
@BrianM.Scott and that suffices to prove its a field? – Sam Harper Oct 18 '16 at 13:36
@Sam: All you need is multiplicative inverses, right? An integral domain meets all of the other requirements to be a field. – Brian M. Scott Oct 18 '16 at 13:37

Bernard · Answer 2 · 2016-10-18T14:53:23.740

0

Hint:

If $R$ is an integral domain, multiplication by $a\ne 0$ in $R$ is an injective ring homomorphism. Now, for a map between sets with the same finite cardinality, $$\text{injective}\iff\text{surjective}\iff\text{bijective}. $$

edited Oct 18 '16 at 14:53

answered Oct 18 '16 at 13:24

Bernard

175,478

Why this down vote? – Bernard Oct 18 '16 at 14:51

Prove R is a Field

2 Answers2