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 $\displaystyle \bar{R} = R - {0} $. Now since R is finite, so is $\displaystyle \bar{R}$. Since $\displaystyle \bar{R}$ is finite and closed under multiplication, take some element $\displaystyle x \in \bar{R} $ and $\displaystyle x * \bar{R} = \bar{R} = \bar{R} * x $ (This is because no zero divisors imply cancellation laws, so $\displaystyle x*x_0 = x*x_1 \text{ means } x_0 = x_1$. so There exists a $\displaystyle x_0 \in \bar{R} \text{ such that } x*x_0 = x $. Now for any element $\displaystyle y \in \bar{R} $ $\displaystyle x*x_0*y = x*y $ Since, no zero divisors imply cancellation laws are present, $\displaystyle x(x_0*y)=x(y) \text{ means } x_0*y = y $. Since $\displaystyle y * \bar{R} = \bar{R} $ say $\displaystyle y*x_1 = y $. So $\displaystyle x_0*y = y*x_1 = y $ Multiply both right side by y, you get $\displaystyle x_0*y*y = y*x_1*y = y^2 = y*(x_1*y) $ which means $\displaystyle y = x_1*y = y*x_1 $. Since y was any element from $\displaystyle \bar{R} $. the element $\displaystyle x_1 $ is the identity.

As for Inverses, since for any $\displaystyle x \in x*\bar{R} = \bar{R} =\bar{R}*x $ there exists some $\displaystyle x_0 \in R $ such that $\displaystyle x * x_0 = 1 = x_1 * x $. multiply left by$\displaystyle x_1$, you get $\displaystyle x_1*x*x_0 = x_1 $ and so since $\displaystyle x_1*x = 1 $, we get $\displaystyle 1*x_0 = x_1 $, so $\displaystyle x_0 * x = x * x_0 = 1 $ thus x has an inverse in $\displaystyle \bar{R} $.

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 $\displaystyle \bar{R} = R - {0} $. Now since R is finite, so is $\displaystyle \bar{R}$. Since $\displaystyle \bar{R}$ is finite and closed under multiplication, take some element $\displaystyle x \in \bar{R} $ and $\displaystyle x * \bar{R} = \bar{R} = \bar{R} * x $ (This is because no zero divisors imply cancellation laws, so $\displaystyle x*x_0 = x*x_1 \text{ means } x_0 = x_1$. so There exists a $\displaystyle x_0 \in \bar{R} \text{ such that } x*x_0 = x $. Now for any element $\displaystyle y \in \bar{R} $ $\displaystyle x*x_0*y = x*y $ Since, no zero divisors imply cancellation laws are present, $\displaystyle x(x_0*y)=x(y) \text{ means } x_0*y = y $. Since $\displaystyle y * \bar{R} = \bar{R} $ say $\displaystyle y*x_1 = y $. So $\displaystyle x_0*y = y*x_1 = y $ Multiply both right side by y, you get $\displaystyle x_0*y*y = y*x_1*y = y^2 = y*(x_1*y) $ which means $\displaystyle y = x_1*y = y*x_1 $. Since y was any element from $\displaystyle \bar{R} $. the element $\displaystyle x_1 $ is the identity.

As for Inverses, since for any $\displaystyle x \in x*\bar{R} = \bar{R} =\bar{R}*x $ there exists some $\displaystyle x_0 \in R $ such that $\displaystyle x * x_0 = 1 = x_1 * x $. multiply left by$\displaystyle x_1$, you get $\displaystyle x_1*x*x_0 = x_1 $ and so since $\displaystyle x_1*x = 1 $, we get $\displaystyle 1*x_0 = x_1 $, so $\displaystyle x_0 * x = x * x_0 = 1 $ thus x has an inverse in $\displaystyle \bar{R} $.

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?

Quote:

Originally Posted by

**Rita** jakncoke

Excuse me, Jakncoke, I have one question.

[IMG]http://latex.codecogs.com/png.latex? x * \bar{R} = \bar{R} = \bar{R} * x[/IMG] (This is because no zero divisors imply cancellation laws, so [IMG]http://latex.codecogs.com/png.latex?x*x_0 = x*x_1 \text{ means } x_0 = x_1[/IMG].)

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

Basically, if $\displaystyle xR \not = R $ since R is closed undermultiplication then it means atleast some 2 elements collapsed into the same element, meaning for some $\displaystyle x_0, x_1 \in R $ $\displaystyle x*x_0 = x*x_1 $, since the cancellation law says if $\displaystyle ab = ac \text{ then b = c } $, we have since $\displaystyle x*x_0 = x*x_1 $ then $\displaystyle x_0 = x_1 $, which would mean that two elements didnt collapse into one element after multiplication by x. Thus, every $\displaystyle x*x_i \text { where } x_i \in R $ returns a unique element in the ring R and thus we get the original ring back even after multiplication by x.

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!