If $e^2=e$, then $(e+(1-e)re)$ is an idempotent.

Question

Suppose that in a unital ring $R$, that $e^2=e$, then I want to show that $(e+(1-e)re)$ is an idempotent (that it's square is equal to itself) for any $ r \in R$.

Attempt:

$$(e+(1-r)re)^2=(e+(1-e)re)(e+(1-e)re) =(ee+e(1-e)re+(1-e)ree+(1-e)re(1-e)(re))$$

And now noting that $(e)(1-e)=(e-e^2)=0$,

$$=(e+(1-e)re)$$

Edit: The original problem had a typo, where I tried to show that $e+(1-r)(re)$ was an idempotent, which is false.

Answer 1

It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.

But suppose we look at $e + (1 - e)re$; we have

$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$ $= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! \tag 1$

So $e + (1 - e)re$ is in fact idempotent; we have used

$e^2 = e \tag 2$

in deriving (2), since it is equivalent to

$(1 - e)e = e(1 - e) = e - e^2 = 0. \tag 3$

I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will

$e + (1 - e)re \ne e? \tag 4$

well, if and only if

$(1 - e)re \ne 0, \tag 5$

which is to say

$re \ne ere; \tag 6$

so if we want to generate new idempotents from old, we need find those $r \in R$ such that (6) binds.

Answer 2

The claim is not true in $\mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have

$$ e+(1-r)re = 1 + r -r^2 $$ and the polynomial on the RHS clearly doesn't take only idempotents as values.

Answer 3

I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then $$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$