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Prove that every finite domain contains an identity element.

Please give me help

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Consider $a$, $a^2$, $a^3$, and so on. By finiteness, there exist natural numbers $m<n$ such that $a^m=a^n$. – André Nicolas Mar 9 '12 at 16:26
What exactly do you mean by domain? It sounds like you are asking how to show that every finite integral domain is a field, is that maybe what you wanted to say? If so, this is just Wedderburn's little theorem. – Lepidopterist Mar 9 '12 at 16:27
Please don't replace titles with nonsense. If you meant "field" instead of domain, then fields, by definition have both a multiplicative and an additive identity, whether they be finite or infinite. – Arturo Magidin Mar 9 '12 at 21:19
@jay: I assume the question is to show that a finite ring (not necessarily with a unit) which has no zero divisors must in fact have a unit. – Qiaochu Yuan Mar 9 '12 at 22:02

Let $a$ be a non-zero element of the domain. Consider the objects $a$, $a^2$, $a^3$, and so on.

The powers of $a$ cannot all be different, since if they were, there would be infinitely many elements in the domain. It follows that there are natural numbers $m$ and $n$, with $m<n$, such that $a^m=a^n$. Thus for any $x$ in the domain, $$xa^n=xa^m,$$ and therefore $$xa^{n-m}a^m=xa^m. \qquad(\ast)$$ Since $a\ne 0$, we have $a^m\ne 0$, so by $(\ast)$ and the cancellation property, $$xa^{n-m}=x.$$ This says that $a^{n-m}$ is a multiplicative identity.

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Hint $\:$ The elements $\ne 0$ form a nonempty finite semigroup, so contain an idempotent $\rm\:e.\:$ Thus $$\rm\: 0\: \ne\: e\: =\: e^2\: \ \Rightarrow\ \ e\:x\: =\: e^2\:x\ \ \Rightarrow\ \ x\: =\: e\:x$$

Alternatively, notice that $\rm\ a\ne 0\:\Rightarrow\: x\mapsto a\:x\ $ is $1$-$1$ so onto, therefore

$\qquad\qquad$ for all $\rm\:x\!:\ $ $\rm\begin{eqnarray}\exists\: e\!:\ \ a\: &=&\:\rm \color{#C00}{a\:e}\\ \rm \exists\: d\!:\ \ x\: &=&\:\rm a\:d\end{eqnarray}$ $\ \ \Rightarrow \ \ \begin{eqnarray}\rm a\:d \: &=&\: \rm \color{#C00}{e\:a}\:d\\ \rm x \:&=&\: \rm e\:x\end{eqnarray} $

The proof in André's answer is a special case of the first method above. Probably the proof of Herstein mentioned by Chandrasekhar is similar to the second proof above.

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This is a property that every finite integral domain $D$ is a field. Please look into I.N.Herstein's text.

Idea. Is to consider a non-zero element in $D$ and establish a bijection from $D \to D$ via $x \mapsto ax$.

Please look into Page 128 of the following link

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thank you so much for your reply, i have got each finite integral domain is a field noww. so field contains identity element?? – zile xu Mar 9 '12 at 17:24
@zilexu: Yes, see the definition of a field. – user9413 Mar 9 '12 at 18:09

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