Let $\mathcal{A}$ be a $K$-algebra and $e_1,e_2 \in \mathcal{A}$ be idempotents such that $e_1 e_2 =e_2 e_1$ and $e_1 - e_2\in \mathcal{J}(\mathcal{A})$ (the Jacobson Radical of $\mathcal{A}$). Show that $e_1 = e_2$.
Help please, I think i already tried everything :(
TIP: $\forall a\in \mathcal{J}(\mathcal{A})$ we have that $1+xa$ is a unity (has inverse) in $\mathcal{A}$ for all $x\in\mathcal{A}$
That's why I'm trying to use.
You want to show $e_1 - e_2 = 0$ so try muptiplying $e_1 - e_2$ by an invertible element of the form $1 + a(e_1 - e_2)$. To get an idea of which $a$ to use, we multiply $(e_1 - e_2)$ on the left by $(e_1 - e_2)$.
So $$(e_1 - e_2)^2 = e_1 - 2e_1e_2 + e_2.$$ If we take $a = -e_1$ then we have $$e_1(e_1 - e_2)^2 = e_1 - 2e_1e_2 + e_1 e_2 = e_1 - e_1e_2$$ and hence,
$$ \big( 1 - e_1(e_1 - e_2) \big)(e_1 - e_2) = e_1 - e_2 - e_1 + e_1e_2 = e_1e_2 - e_2. $$
Again, multiply on the left by $e_1 - e_2$ to look for a candidate for $a$: $$(e_1 - e_2)(e_1e_2 - e_2) = -(e_1e_2 - e_2).$$ So
$$ \big( 1 + (e_1 - e_2) \big)\big( 1 - e_1(e_1 - e_2) \big)(e_1 - e_2) = \big( 1 + (e_1 - e_2) \big)(e_1e_2 - e_2) = 0 $$
and it follows that $e_1 = e_2$.