A ring becomes a division ring?

**Rita** · December 17th 2012, 10:36 PM

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?

**jakncoke** · December 18th 2012, 12:25 AM

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

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

**Deveno** · December 18th 2012, 03:55 AM

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: a² + ab = a², 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).

**Rita** · December 18th 2012, 10:31 AM

?

**Rita** · December 18th 2012, 10:37 AM

Originally Posted by jakncoke

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

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

jakncoke
Excuse me, Jakncoke, I have one question.
"xR=R=Rx(This is because no zero divisors imply cancellation laws, so xx₀=xx₁implies x₀=x₁.)" (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?

**jakncoke** · December 18th 2012, 10:38 AM

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

**Deveno** · December 18th 2012, 10:38 AM

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)

**Rita** · December 23rd 2012, 05:50 PM

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

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