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

1. ## 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!

2. ## 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 $Re + Rf = R = R(e+f)$

3. ## 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. ??

4. ## 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

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

Originally Posted by jakncoke
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.