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Question: Let R be a finite ring without zero divisors and |R| > 1. Then show that R is a division ring.

I used the following logic, but I'm not sure if this is correct.

Let |R| = n, a finite number. $Let 0 \not = a \in R.$

Therefore, $a^n = a => a(a^{n-1} - 1) = 0 => a^{n-1} = 1 \,\,\, (\text{Since} \,\, a \not = 0)$

Therefore, $a a^{n-2} = 1$

Therefore $a^{n-2}$ is the inverse of a for all a. Therefore R is a division ring.

Does this work?

Doodle Bob
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user1742188
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2 Answers2

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Why do you think that $a^n=a$? This is not obvious at all. It will end up being true, because if it is a division ring then the nonzero elements will form a group under multiplication, and then it follows from Lagrange's theorem. But you have to prove it first.

First of all, by induction it follows that if $a\ne 0$ then $a^n\ne 0$ for all $n\in\mathbb{N}$, this is because there are no zero divisors. Now, it is true that there are some $j>i$ such that $a^j=a^i$, because the set $\{a,a^2,a^3,a^4,...\}$ must be finite. Then $0=a^j-a^i=a^i(a^{j-i}-1)$. Since $a^i\ne 0$ it follows that $a^{j-i}-1=0$, again because there are no zero divisors. So now $a^{j-i-1}$ is an inverse of $a$.

Mark
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For any

$0 \ne a \in R \tag 1$

consider the map

$\theta_a: R \to R \tag 2$

given by

$\theta_a(r) = ar, \; \forall r \in R; \tag 3$

if

$\theta_a(r) = \theta_a(s), \tag 4$

then

$ar = as, \tag 5$

that is,

$a(r - s) = 0; \tag 6$

since $R$ has no zero divisors, by virtue of (1) we conclude that

$r - s = 0, \tag 7$

whence

$r = s, \tag 8$

and we see that $\theta_a$ is injective; $R$ being finite, we also have $\theta_a$ is surjective, and therefore

$\exists b \in R, \; ab = \theta_a(b) = 1, \tag 9$

and $b$ is a right inverse of $a$; now

$ab = 1 \Longrightarrow (ba)b = b(ab) = b \cdot 1 = b, \tag{10}$

from which

$(ba -1)b = (ba)b - b = 0, \tag{11}$

and once more invoking the hypothesis that $R$ has no zero divisors (and observing that (9) implies $b \ne 0$) we may write

$ba - 1 = 0 \Longrightarrow ba = 1 \tag{12}$

as well; we see that $b$ is also a left inverse of $a$; (9) and (12) together imply that

$b = a^{-1}; \tag{13}$

since every non-zero element $a$ of $R$ has a (two-sided) inverse, $R$ is indeed a division ring. $OE\Delta.$

Robert Lewis
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