@Peter

Let we ring  have some idempotent and  some invertible element( not a field or special ring)

If we have idempotent element can we find invertible element??

Or converse??

@Peter

Let me 

Suppose E be an idempotent

Is E^2 or xE or 1-E is a unit??

I know they are not units but may be a example such them be a unit

There is only one idempotent which is a unit: 1.

This comes from e(1-e)=0, that is, an idempotent is a zero divisor. This also tells you that is a field there are only two idempotents: 0 and 1.

i know what happen for E and 1-E

i use them for example what i want in really

maybe you can thinking about linear combinition or product of n elements or other way

an idempotent in a unitary ring with zero divisors is a zero divisor as said Pedro, so there is no relation between it and units (divisors of one=invertible elments).

In an integral ring, we have only 1 and 0.

Numerical example: in Zn,  the arithmetic modular ring with n not prime, the idempotent e is that  which satisfies  e(1-e)  is a multiple of n, while the units are those co-prime to n.

@Hanifa

what you can say for ring with no zero divisor??  

a unitary commutative ring with non zero divisors is called an integral domain. This one has no idempotent only 1 and zero. 

Examples: Z, Zp for p prime, any polynomial ring on integral domain, any primitive extension of an integral ring ( for example Z[ square(α)],  α in R or C ), many many examples of integral domains.

what is the relation between idempotent and involution elements?

thanks

but you thinks we   never  can compute  a unit element if we have an idempotent???

I think matrix ring, i.e. the set of matrices over some ring R under matrix addition and multiplication, gives a good example;

unit matrix is idempotent: I2 = I,

unit matrix is invertible: I -1= I.

 @Tadeusz

Thank you 

Thats right 

Can you think abstractly about all ring??

Let R be a ring with unity 1R and suppose R has no zero divisor.The only idempotent in R are 0R and 1R.

Clearly 0R and 1R are idempotent elements. Let e ∈ R be an idempotent. Then e2 = e and so e(e -1R) = 0R. Since R has no zero divisor, either e = 0R or e - 1R = 0R, i.e., either e = 0R or e = 1R. Therefore the only idempotents of R are 0R and 1R.

Dear Yaser,

For the above question "what is the relation between idempotent and involution elements?"

If a is an idempotent of the endomorphism ring End_{R}(M), then the endomorphism f = 1-2a is an R module involution of M. That is, f is an R homomorphism such that f² is the identity endomorphism of M.

Please see the following useful link:

https://en.wikipedia.org/wiki/Idempotent_element

Regards,

Wiwat Wanicharpichat

To address all questions you have been asking in the thread:

1) A zero divisor is never an invertible element: Suppose otherwise that we have ab=0 with a,b not equal to zero and a invertible. Then there exists an element c such that ca=1. But then b=1b = (ca)b = c(ab) = c0=0. Contradiction!

2) In a ring with identity, any idempotent element is either 0, 1 or a zero divisor: since e^2=e we get e^2-e=0 or e(1-e)=0. Now if e is not 0 or 1, then e and 1-e are both nonzero and hence zero divisors.

3) Mixing 1) and 2) we get that no idempotent except 1 can be an invertible element.

4) As Wiwat Wanicharpichat and Peter Breuer have mentioned, if e is an idempotent then the element a:=1-2e is a square root of unity, since (1-2e)(1-2e) = 1-4e+4e^2=1-4e+4e=1, so a^2=1 and a is invertible with a^(-1)=a.

Moreover, if 2 is invertible in your ring, then from every square root of unity a you get an idempotent e:=(1-a)/2, since e^2=(1-a)^2/4 = (1-2a+a^2)/4 = (1-2a+1)/4 =(2-2a)/4=(1-a)/2=e (note that under this correspondence, e=0 when a=1 and e=1 when a=-1).

5) You can also get invertible elements from nilpotent ones: for example, if a^2=0, then (1-a)(1+a) = 1-a^2 = 1 and thus 1+a is the inverse of 1-a.

EDITED 03/09/18:

6) More in general: Suppose you have an algebraic element x in your ring R, which satisfies the univariate polynomial P over some field (e.g. the rational field), and let Q be another univariate polynomial which is relatively prime to P, i.e., gcd(P,Q)=1. Then, by Bézout's identity for polynomials, there are polynomials A,B such that A(X)P(X)+B(X)Q(X)=1. Now evaluation by x from R gives A(x)P(x)+B(x)Q(x)=1, so B(x)Q(x)=1 since P(x)=0. Therefore Q(x) is invertible with B(x)=Q(x)^(-1), so long as the required scalars from the field are present in R. On the other hand, if gcd(P,Q)=D, then Q(x)=D(x)Q'(x) cannot be invertible, since P(x)=D(x)P'(x) implies that P'(x)Q(x)=P'(x)D(x)Q'(x)=P(x)Q'(x)=0, so Q(x) is either 0 or a zero divisor (which cannot be invertible).

We recover the previous examples as particular cases:

a) If x is an idempotent, x^2=x, then P(X)=X^2-X=X(X-1), so every element of the form Q(x) with gcd(Q(X),X(X-1))=1 is invertible in R (if the right scalars are in R). In particular this is true for Q(X)=-2X+1.

b) If x is nilpotent, x^k=0, then P(X)=X^k, so every element of the form Q(x) with Q(X) having nonzero independent term is invertible in R (if the right scalars are in R). In particular this is true for Q(X)=X+1.

@jose

What about linear ornon-linear  combination of idempotents??

Oh dear Yaser the last question is not simple, it can be a search :)

For example, in Linear Algebra, projectors are idempotents and there are many papers on  linear combinations  of them, which of course not an idempotent only under conditions and adding information like orthogonality.

See for example:

J.K. Baksalary and O.M. Baksalary, Idempotency of linear combinations of two idempotent matrices. Linear Algebra Appl. 321 (2000), 3-7.

H. Ozdemir and A.Y. Ozban, On idempotency of linear combinations of idempotent matrices. Applied Mathematics and Computation. 159 (2004), 439-448.

No idempotent except 1 can be an invertible element. In a ring with identity, any idempotent element is either 0, 1 or a zero divisor: since e^2=e we get e^2-e=0 or e(1-e)=0. Now if e is not 0 or 1, then e and 1-e are both nonzero and hence zero divisors.

the element which is idempotent and invertible in a ring is unity (if ring is with unity).

Example : If e is an idempotent then 2e-1 is invertible element 

@Peter

(2e-1)*(2e-1)=1

Yaser, it is as if Wiwat has said the same as you, but with other words. 

André: what you have proved is that there are not invertible idempotents EXCEPT the unity 1. Ie if e² = e and e is a unit then e=1.

And I pointed out that you have to exclude e=1. 1 is an idempotent which is not a zero divisor. Nor e=0.

A trivial answer in a commutative ring.

A unit element a is idempotent iff it equals 1.

An idempotent e is unit iff it equals 1.

Consider the ring M_2(R^+) of square matrices of non negative real number. Then the 2x2 matrix (1/2 1/2 1/2 1/2) is idempotent but is of rank 1 and not invertible.

- If e is an idempotent, then 1-2e is invertible.

- If e is idempotent and u isn unit, then ue^{-1} is is idempotet