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$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?

I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me.

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By "zero factors" you meant "non-trivial zero divisors"? But even if you did the claim is false, as the non-field ring $\,\Bbb Z\,$ proves. – DonAntonio Feb 26 '13 at 15:21
@DonAntonio I assume by finitive he meant finite. – JSchlather Feb 26 '13 at 15:22
Well, that may be so, @JSchlather... then he want to prove the claim that a finite integral domain is a field – DonAntonio Feb 26 '13 at 15:37
up vote 11 down vote accepted

It follows from the pigeon-hole principle.

First, we show that $R$ has an identity. (Sometimes the existence of an identity is included in the definition of a ring, but you do not need it here.)

For each $0 \ne a \in R$, consider the map $$ \phi_{a} : R \to R, \qquad x \mapsto a x. $$ Because $a$ is not a zero-divisor, the map $\phi_{a}$ is injective, thus surjective. So there is $e \in R$ such that $$a = \phi_{a}(e) = e a.\tag{1}$$ Again because the map $\phi_{a}$ is surjective, every $b \in R$ is of the form $$b = \phi_{a}(x_{b}) = a x_{b}$$ for some $x_{b}$, so multiplying (1) by $x_{b}$ you find $$ b = e b $$ for all $b \in R$, so $e$ is the identity.

Now use once more the fact that $\phi_{a}$ is surjective, to show that there is $c \in R$ such that $$a c = \phi_{a}(c) = e,$$ so $c$ is an inverse of $a$, $a$ being an arbitrary non-zero element.

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This is a classic exercise. It's more precisely stated as a finite integral domain with more than one element is a field.

Here's a hint. Let $a \in R$ be non-zero and consider the map $\phi: R \rightarrow R$ given by $\phi(r)=ra$. What can you say about $R$? Is it injective? What's in the image?

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Another possibility: Let $x\in R$, $x\neq 0$. Since $R$ is finite, there must be some repetition in the infinite list $x^0, x^1, x^2, x^3, x^4,\ldots$. Hence there are exponents $i < j$ with $x^i = x^j$. So $x^i(x^{j-i} - 1) = 0$. Since $R$ has no zero divisors, we get $x^{j-i} - 1 = 0$ ($x^i \neq 0$ since $x\neq 0$ and $R$ has no zero divisors). So $x\cdot x^{j-i-1} = 1$, showing that $x$ is invertible.

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I.e. $\rm\:x \to ax\:$ is an invertible map on a finite set, so is a permutation, so has finite order. – Math Gems Feb 28 '13 at 2:10

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