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How can I prove this:

Finite rings without zero divisors are division rings.

I know how to prove it when the ring has $1$, but I have no idea if my ring needs to have an unity.

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  • $\begingroup$ Let $r$ be an arbitrary element and consider the sequence $\{r,r^2,r^3,\dots\}$ $\endgroup$ – Max Dec 16 '16 at 18:55
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Since this statement of the question clearly goes haywire if we allow $\{0\}$ to be admitted, I'll be working under the assumption that the ring has at least one element other than zero. I guess also that "without zero divisors" is meant to eliminate both left and right zero divisors. (Update: it occurred to me later how to show a ring with no nonzero zero divisors has no nonzero one sided zero divisors. If $ab=0$ for nonzero $a,b$, then $ba\neq 0$ and $(ba)^2=0$, a contradiction. So, no one sided zero divisors exist. )

Notice multiplication by nonzero elements must permute the finite set of nonzero elements of the ring. Let $a$ be any nonzero element. There exists a $b$ such that $ab=a$. Automatically, $aba=a^2$ implies $ba=a$ as well. $b$ is our candidate for identity.

Now let $c$ be arbitrary and nonzero. There exists a $d$ such that $da=c$. But then $c=da=dab=cb$ implies $c=cb$ and again this leads to $c^2=cbc$and $bc=c$.

Thus $b$ acts as an identity for arbitrary nonzero elements. It trivially acts as an identity for $0$. Thus $b$ is the identity for the ring.


Here's another amusing proof. Call this ring $R$ and look at the Dorroh extension $\Bbb Z\times R=\{(z,r)\mid z\in \Bbb Z, r\in R\}$. In case you're not familiar with it, addition is $(z,r)+(z',r')=(z+z',r+r')$ and multiplication is $(z,r)(z',r')=(zz',zr'+z'r+rr')$. It creates a new ring $R_1$ which has $R$ as an ideal.

Since $R^2=R$ and $R$ is finitely generated, we can apply Nakayama's lemma to $R$ to get the existence of an element $x\in R$ such that $(1-x)R=\{0\}$. Unpacking that, we see that $x$ is a left identity element of $R$. Doing the same thing on the right, there is a right identity (necessarily also $x$.)

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  • $\begingroup$ Why there is such $b$ that $ab = a$? $\endgroup$ – Yola Sep 15 '17 at 8:16
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    $\begingroup$ @Yola Notice multiplication by nonzero elements must permute the finite set of nonzero elements of the ring. The permutation caused by $a$ must map something to $a$. $\endgroup$ – rschwieb Sep 15 '17 at 10:36
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Consider, for $a\in R$, the map $r_a\colon R\to R$ defined by $r_a(x)=xa$; this map is injective if $a\ne0$, hence surjective. So there's an element $b$ such that $ba=a$. Of course $b\ne0$; therefore there exists $c\in R$ such that $bc=b$. Then, for each $x\in R$, $$ bx=bcx $$ and so $x=cx$. Therefore there is a left identity. Similarly there's a right identity.

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  • $\begingroup$ @rschwieb I fixed the argument $\endgroup$ – egreg Dec 4 '13 at 7:28
  • $\begingroup$ Excellent! All ok now. $\endgroup$ – rschwieb Dec 4 '13 at 10:58

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