text: # A ring becomes a division ring?
  
  • Dec 17th 2012, 10:36 PM
  Rita
  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?
  • Dec 18th 2012, 12:25 AM
  jakncoke
  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 $\displaystyle \bar{R}$.
  • Dec 18th 2012, 03:55 AM
  Deveno
  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: 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).
  • Dec 18th 2012, 10:31 AM
  Rita
  Re: A ring becomes a division ring?
  ?
  • Dec 18th 2012, 10:37 AM
  Rita
  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 xx0=xx1 implies 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?
  • Dec 18th 2012, 10:38 AM
  jakncoke
  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.
  • Dec 18th 2012, 10:38 AM
  Deveno
  Re: A ring becomes a division ring?
  suppose we have cancellation.
  
  then Lx:R*-->R* given by Lx(r) = rx is an injective function:
  
  if xr = xr', then r = r'.
  
  this, in turns means Lx is a surjective function, since R is finite.
  
  that is: xR* = R* (xR is the image Lx(R)).
  
  (this only works for non-zero x)
  • Dec 23rd 2012, 05:50 PM
  Rita
  Re: A ring becomes a division ring?
  Big thanks to both of you, and, Gosh, you're so smart!