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I'm attempting to understand the proof of the statement that every finite domain $R$ is a division ring. For a fixed nonzero element $a \in R$, the left-translation map is injective since $R$ is a domain. Since $R$ is finite this map is also surjective which implies that 1 is in its image. Therefore we have the existence of an element $b \in R$ such that $ab=1$, right inverses exist for nonzero elements. By looking at the right-translation map, we get the existence of left inverses.

Given a finite domain $R$ and $a,b,c \in R$ such that $ab=1$ and $ca=1$ how do we show that $b=c=a^{-1}$? How do we know that the nonzero elements are units?

I think the answer is as follows: The nonzero elements already form a multiplicative monoid since $R$ is a domain. But to form a group it is enough to demonstrate the existence of left (or right) inverses only by these alternate group axioms.

So maybe we just need left inverses and an identity to get a division ring.

Bill Dubuque
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dls
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  • I notice that you have not accepted any of the answers given in any of the three questions you asked. Remember to "accept" the answer you find most helpful if there is such a one, by clicking on the check mark to the left of the question. See the last paragraph under How do I ask questions here? in the FAQ, http://math.stackexchange.com/faq – Arturo Magidin Dec 29 '10 at 22:29
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    Did not realize the distinction between up-voting and accepting. Thanks. – dls Dec 29 '10 at 22:37
  • You can upvote any or all answers (or any subset thereof), whether or not you asked the question. Only the person asking the question can accept an answer, and only one answer (though you can change your mind and "un-accept" an answer later, or decide to accept a different one). – Arturo Magidin Dec 29 '10 at 22:38

4 Answers4

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In any ring $R$, if $a$ has both a left and a right inverse, then the left and right inverses are the same and the element is a unit.

To see this in your case, just write this: $$ b = 1\cdot b = (ca)b = c(ab) = c\cdot 1 = c.$$ Since $b=c$, then $ca = ba = ab = 1$, so $b=c=a^{-1}$, as desired.

This is the standard argument. It can be used, for example, to show that a function that has left and right inverses must be invertible (and the left and right inverses are the same).

It is also true that if you have a ring without zero divisors, which has a left identity $e\neq 0$ and every element has a left inverse, then the ring is a division ring. The argument is the one you give, essentially showing that the nonzero elements form a group under multiplication. But the fact that an element having both left and right inverses is in fact invertible is more basic than that.

Arturo Magidin
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  • Thanks, I wasn't seeing that! – dls Dec 29 '10 at 22:22
  • @dls: You'll see it many times over; it does show up a lot. (-: – Arturo Magidin Dec 29 '10 at 22:23
  • I think my reasoning was correct, too, being linked together by this article: http://groupprops.subwiki.org/wiki/Equality_of_left_and_right_inverses_in_monoid – dls Dec 29 '10 at 22:27
  • @dls: See my last paragraph; yes, you can argue the way you did that the nonzero elements form a group; in fact, they form a group of $S_{R-{0}}$, the group of permutations of the nonzero elements of $R$ under the map that sends $r$ to the permutation obtained by left-multiplication-by-$R$. But the argument Matt and I give is the very basic and very standard one, and it does not require you to assume quite so much in order to show that $b=c$. You would have a hard time making your argument work if the ring was not a domain, though it's true, if you restrict yourself to the nonzero divisors. – Arturo Magidin Dec 29 '10 at 22:31
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I thouht I would add an extra answer that goes in a slightly different direction than the rest. An alternative proof that a finite domain is a division ring goes as follows: Look at different powers of $a$ (ie, $a^n$ for various positive numbers $n$). Now, since the ring is finite, you will at some point get to some element you have already been at before, so you have $a^n = a^m$ with $n < m$. Now we can use that it is a domain, so this leads to $a^{m-n} = 1$ which means that not only does $a$ have an inverse. his inverse is in fact a power of $a$ (and then clearly it commutes with $a$ so it is also an inverse from the other side).

This argument is less elegant than the usual one, but I find it is nice to know that the inverse of an element in a finite ring is actually a power of the element itself.

Tobias Kildetoft
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What happens if you mutliply the identity $ab = 1$ on the left by $c$?

Matt E
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Hint $ $ Unify the equations $\rm\ ca = 1,\,\ ab = 1\ $ by creating a common generalization of their LHS,$\ $ namely $\rm\ cab.\: $ Now apply the two equations to the unification $\rm\ cab\ $ to deduce that $\rm\ c = b,\,$ i.e.

$$\rm c = c(ab) = (ca)b = b\qquad\qquad\qquad$$

Said more concisely: $ $ simply evaluate $\rm\ \overline{ c\, } \overline{ \underline {a}} \underline{\,b}\ $ in $\,2\,$ ways, using each over/underlined term $ = 1$.

This uniqueness of inverses is used in many places, e.g. a common proof of the Law of Signs.

Remark $ $ A similar procedure often works to generate consequences of equations considered as normal-form rewrite rules. This may be used to automatically derive normal-form rewriting algorithms for various algebraic structures, e.g. look up the Knuth-Bendix completion algorithm. Many well-known algorithms may be considered as special cases of such, e.g. Gaussian elimination and its nonlinear generalizations such as the Grobner basis algorithm, and also the Todd-Coxeter algorithm for enumerating cosets of group presented by generators and relations.

See also this post on the uniqueness of roots of $\,\rm ax = b$.

Bill Dubuque
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