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!
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!
Notice if R is regular, then such that and so Now how do we know that x isn't a zero divisor? and thus ? Well, if then and so which means , which cannot be so, and thus, , so , 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, and , so basically both
because i wanted the result that so which would mean x was a unit. Now if you remember, if is a zero divisor then a element such that but neither a nor b is 0. So if then i cannot get the result i want, which is that x is a unit
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 =
, we may take y =
.
otherwise we have x = PAQ, where A is the matrix:
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:
,
and set y = Q^{-1}BP^{-1}, we have:
x = PAQ = P(ABA)Q (from above, A = ABA)
= PA(QQ^{-1})B(P^{-1}P)AQ
= (PAQ)(Q^{-1}BP^{-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.