A ring becomes a division ring?

I got one problem from the textbook:

Prove that a finite ring with more than one element and no zero divisors is a division ring.

It's obvious that the zero element must exist in this ring, so "with more than one element" means this ring is not trivial.

And once I prove that the identity element 1 exists in this ring, then the rest of work is easy.

So my question is reduced to the existence of the identity element.

Any hint?

Re: A ring becomes a division ring?

Ok. Let R be the ring. Since R has more than one element, take . Now since R is finite, so is . Since is finite and closed under multiplication, take some element and (This is because no zero divisors imply cancellation laws, so . so There exists a . Now for any element Since, no zero divisors imply cancellation laws are present, . Since say . So Multiply both right side by y, you get which means . Since y was any element from . the element is the identity.

As for Inverses, since for any there exists some such that . multiply left by , you get and so since , we get , so thus x has an inverse in .

Re: A ring becomes a division ring?

the problem here is exactly the same as:

any finite (non-empty) semi-group S with the cancellation property is a group.

left cancellation means that left-multiplcation is injective (similarly for right-cancellation). on a finite set endomorphism (mapping of a set to itself), this implies bijectivity (the finiteness condition is necessary, here, as the "doubling function" is injective on the integers, but not surjective).

but this means, given any a in S, we have some (particular) x with a*x = a, since a*__ is surjective, and since __*a is likewise surjective, for any b in S, we have b = u*a.

thus for that particular x, and ANY b: b*x = (u*a)*x = u*(a*x) = u*a = b <---note how associativity of * is necessary, here.

now x has only been show to be a "right-identity", so far, but a similar argument shows we have a "left-identity" y, whence:

y = y*x = x. we can call x = y, e, now.

of course, now, we have some element a' such that a*a' = e (since e is in S and a*__ is surjective).

since (a'*a)*a' = a'*(a*a') = a'*e = e*a', (right) cancellation now implies a'*a = e, so every element possesses a (two-sided) inverse, which must be unique.

if our semi-group S, is the non-zero elements of a ring R, and this set is non-empty, cancellation is the same thing as "no-zero divisors":

suppose R -{0} has no zero divisors, and let a,b,c be in R - {0}.

then if ab = ac, ab - ac = 0, so a(b - c) = 0. since a ≠ 0, we must have b - c = 0, so b = c. a similar proof holds for ac = bc.

on the other hand, suppose we have cancellation:

then if ab = 0, and a ≠ 0, we have: a^{2} + ab = a^{2}, that is: a(a + b) = a(a), so a + b = a, so (subtracting a from both sides) b = 0 (i had to think for a bit to find a form that did not assume R has unity).

thus (left) cancellation means no (left) zero divisors and similarly for right cancellation and right zero divisors.

a more interesting result (and somewhat harder to prove is):

such a ring is commutative (a finite division ring is a field).

Re: A ring becomes a division ring?

Re: A ring becomes a division ring?

Quote:

Originally Posted by

**jakncoke** Ok. Let R be the ring. Since R has more than one element, take

. Now since R is finite, so is

. Since

is finite and closed under multiplication, take some element

and

(This is because no zero divisors imply cancellation laws, so

. so There exists a

. Now for any element

Since, no zero divisors imply cancellation laws are present,

. Since

say

. So

Multiply both right side by y, you get

which means

. Since y was any element from

. the element

is the identity.

As for Inverses, since for any

there exists some

such that

. multiply left by

, you get

and so since

, we get

, so

thus x has an inverse in

.

jakncoke

Excuse me, Jakncoke, I have one question.

"xR=R=Rx(This is because no zero divisors imply cancellation laws, so xx_{0}=xx_{1 }implies x_{0}=x_{1}.)" (Sorry, I omitted all *.)

Why is that? I mean... I can't imagine the connection between cancellation laws and xR=R. Could you explain it in more detail?

Re: A ring becomes a division ring?

Re: A ring becomes a division ring?

suppose we have cancellation.

then L_{x}:R*-->R* given by L_{x}(r) = rx is an injective function:

if xr = xr', then r = r'.

this, in turns means L_{x} is a surjective function, since R is finite.

that is: xR* = R* (xR is the image L_{x}(R)).

(this only works for non-zero x)

Re: A ring becomes a division ring?

Big thanks to both of you, and, Gosh, you're so smart!