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Math Help - A ring becomes a division ring?

  1. #1
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    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?
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: A ring becomes a division ring?

    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} .
    Last edited by jakncoke; December 18th 2012 at 12:42 AM.
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  3. #3
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    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).
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  4. #4
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    Re: A ring becomes a division ring?

    ?
    Last edited by Rita; December 18th 2012 at 10:37 AM.
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  5. #5
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    Re: A ring becomes a division ring?

    Quote Originally Posted by jakncoke View Post
    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=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?
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: A ring becomes a division ring?

    Quote Originally Posted by Rita View Post
    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.
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  7. #7
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    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)
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  8. #8
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    Re: A ring becomes a division ring?

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