Let $R$ be a ring with unity such that for each $a ∈ R$ there exists $x ∈ R$ such that $a^2x =a$. 
14. Let $R$ be a ring with unity such that for each $a \in R$ there exists $x \in R$ such that $a^2 x = a$. Prove the following:
(e) There exists $y \in R$ such that $a^2 y = a$, $y^2 a = y$, and $ay = ya$.
(f) $aua = a$, where $u = 1 + y - ay$ is invertible, and $y \in R$ is chosen as in (e).

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clearly it exists y such that $a ^ 2y = a$, but I know not how to get that $y ^ 2a = a$ and $ay =ya$ . For part (f) I do not know how to prove that u is invertible. I tried as follows: $(1 + y-ay) (x) = 1$ and came to $yx = ayx$, how much has to be worth x to be given equality?
 A: First prove that $R$ has no nonzero nilpotent elements. Suppose $a\ne 0$ and is nilpotent, i.e. there is a $n>1$ such that $a^n=0,\:a^{n-1}\ne 0$. There exists $x\in R$ such that $a^2x=a$. But $0=a^nx=a^{n-2}a^2x=a^{n-1}\ne0$, which is contradiction. So $R$ has no nonzero nilpotent elements.
Next prove that $axa−a$ is nilpotent, and so $axa=a$. Suppose $axa−a\ne0$. Then
$$
(axa−a)^2=ax\underbrace{aax}_{aax=a}a-a^2xa-axa^2+a^2=axa^2-a^2-axa^2+a^2=0
$$
i.e. $axa−a$ is nilpotent and thus $\:axa-a=0$, i.e $axa=a$.
Next prove that $a=xaa$. 
$$
(a−xaa)^2=a^2-\underbrace{axa}_{axa=a} a-xaa\cdot a+xa\underbrace{axa}_{axa=a}a=-xa^3+xa^3=0
$$
$a−xaa$ is nilpotent and thus $\:a=xaa$.
Next prove that $ax=xa$. 
$$
(ax-xa)^2=axax-axx\underbrace{axa}_{a=a^2x}-x\underbrace{axa}_{aax=a}+x\underbrace{axa}_{axa=a}=axax-ax\underbrace{xaa}_{xaa=a}x-xa+xa=axax-axax=0
$$
$ax-xa$ is nilpotent and thus $\:ax=xa$.
Now let $y=xax$. Then 
$$
a^2y=a^2 xax=aax=a
$$
Moreover 
$$
ay=\underbrace{axa}_{axa=a}x=ax=xa=x\underbrace{axa}_{axa=a}=ya
$$
And
$$
y^2a=xaxx\underbrace{axa}_{axa=a}=xax\underbrace{xa}_{xa=ax}=x\underbrace{axa}_{axa=a}x=xax=y
$$
Finally let $u=1+y-ay$. Then
$$
aua=a(1+y-ay)a=a^2+a\underbrace{ya}_{ay=ya}-\underbrace{a^2y}_{a^2y=a}a=a^2+a^2y-a^2=a
$$
