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Math Help - help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

  1. #1
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    help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    can anyone help me prove this exercisese
    if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f).
    R is regular ring if for every x in R exist y such that x=xyx.

    Thank you!
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    Re: help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    Notice if R is regular, then  \forall x \in R x \not = 0, \exists y such that  xyx = x and so  xyx-x = 0  \to  x(yx-1) = 0 Now how do we know that x isn't a zero divisor? and thus  yx-1 \not = 0 ? Well, if  (yx-1) \not = 0 then  x ( yx - 1 ) \not = 0 and so  xyx - x \not = 0 which means  x-x \not = 0, which cannot be so, and thus,  yx-1 = 0 , so  yx = 1 , which basically means that every non zero element of R is a unit since we picked x at random from R. So, since there exist inverses for f and e,  (Re + Rf)f^{-1} = Ref^{-1} + R = R and  R(e+f)(e+f)^{-1}  =  R , so basically both  Re + Rf = R = R(e+f)
    Last edited by jakncoke; November 9th 2012 at 03:04 PM.
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    Re: help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    did you prove that Re+Rf+R(e+f) ??

    Why you say that (yx-1) diferent from zero, then x(yx-1) different from 0. ??
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    Re: help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    because i wanted the result that  (yx-1) = 0 so  yx = 1 which would mean x was a unit. Now if you remember, if a \in R is a zero divisor then \exists a element  b such that  ab = 0 but neither a nor b is 0. So if yx-1 \not = 0 then i cannot get the result i want, which is that x is a unit
    Last edited by jakncoke; November 10th 2012 at 07:30 AM.
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    Re: help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    Quote Originally Posted by jakncoke View Post
    Notice if R is regular, then  \forall x \in R x \not = 0, \exists y such that  xyx = x and so  xyx-x = 0  \to  x(yx-1) = 0 Now how do we know that x isn't a zero divisor? and thus  yx-1 \not = 0 ? Well, if  (yx-1) \not = 0 then  x ( yx - 1 ) \not = 0 and so  xyx - x \not = 0 which means  x-x \not = 0, which cannot be so, and thus,  yx-1 = 0 , so  yx = 1 , which basically means that every non zero element of R is a unit since we picked x at random from R. So, since there exist inverses for f and e,  (Re + Rf)f^{-1} = Ref^{-1} + R = R and  R(e+f)(e+f)^{-1}  =  R , so basically both  Re + Rf = R = R(e+f)
    we cannot conclude from yx - 1 ≠ 0 that x(yx - 1) ≠ 0. if this were so, every von neumann regular ring would be a field, and this is not so.

    here is a counter-example:

    let R = M(2,Q), the ring of 2x2 matrices over the rationals.

    if a 2x2 rational matrix x is invertible, we may take y = x-1.

    so suppose x is singular. some easy cases, first off:

    if x = 0, any y will do.

    if x =

    \begin{bmatrix}1&a\\0&0 \end{bmatrix}, we may take y =

    \begin{bmatrix}1&0\\0&0 \end{bmatrix}.

    otherwise we have x = PAQ, where A is the matrix:

    \begin{bmatrix}1&a\\0&0 \end{bmatrix}

    and P,Q are invertible matrices (which we get by keeping track of which elementary row/column operations reduce x to A).

    if we label the matrix:

    B = \begin{bmatrix}1&0\\0&0 \end{bmatrix},

    and set y = Q-1BP-1, we have:

    x = PAQ = P(ABA)Q (from above, A = ABA)

    = PA(QQ-1)B(P-1P)AQ

    = (PAQ)(Q-1BP-1)(PAQ) = xyx.

    so this ring is indeed (von neumann) regular, and we have found a non-unit in it (which is also a zero-divisor).

    *********

    i haven't had time to look at this properly, but the fact that e and f are idempotent must be important.
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    Re: help ! if R is regular ring, e and f are idempotents elements then Re+Rf=R(e+f)

    Thank you all for your answers
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