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 $\displaystyle \forall x \in R x \not = 0, \exists y$ such that $\displaystyle xyx = x $ and so $\displaystyle xyx-x = 0 \to x(yx-1) = 0 $ Now how do we know that x isn't a zero divisor? and thus $\displaystyle yx-1 \not = 0$ ? Well, if $\displaystyle (yx-1) \not = 0 $ then $\displaystyle x ( yx - 1 ) \not = 0 $ and so $\displaystyle xyx - x \not = 0$ which means $\displaystyle x-x \not = 0$, which cannot be so, and thus, $\displaystyle yx-1 = 0 $, so $\displaystyle 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, $\displaystyle (Re + Rf)f^{-1} = Ref^{-1} + R = R$ and $\displaystyle R(e+f)(e+f)^{-1} = R $, so basically both $\displaystyle Re + Rf = R = R(e+f) $
because i wanted the result that $\displaystyle (yx-1) = 0 $ so $\displaystyle yx = 1$ which would mean x was a unit. Now if you remember, if $\displaystyle a \in R $ is a zero divisor then $\displaystyle \exists$ a element $\displaystyle b $ such that $\displaystyle ab = 0 $ but neither a nor b is 0. So if $\displaystyle yx-1 \not = 0$ 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 =
$\displaystyle \begin{bmatrix}1&a\\0&0 \end{bmatrix}$, we may take y =
$\displaystyle \begin{bmatrix}1&0\\0&0 \end{bmatrix}$.
otherwise we have x = PAQ, where A is the matrix:
$\displaystyle \begin{bmatrix}1&a\\0&0 \end{bmatrix}$
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:
$\displaystyle B = \begin{bmatrix}1&0\\0&0 \end{bmatrix}$,
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).
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i haven't had time to look at this properly, but the fact that e and f are idempotent must be important.