Prove that the annihilator $A(e)$ of $e$ is a principal ideal

Question

Let $R$ be a commutative ring with $1$ and suppose that $e \in R$ is an idempotent, $e^2=e$.

We need to prove that the annihilator $A(e)$ of $e$ is a principal ideal.

The only thing I can start off with is that $A(e) =$ {$s \in R: es = 0$}.

But I am honestly not sure where to go from here. Any help would be much appreciated.

Answer 1

There is an obvious element of $A(e)$, namely $(1-e)$. Given we have only one suspect, let's try to prove that $A(e)=\langle 1-e\rangle$, i.e., that $es=0$ implies $s=(1-e)t=t-et$ for some $t$. Observe that finding suitable $t$ is not hard: picking $t=s$ works out nicely.

Answer 2

We observe that

$1 = (1 - e) + e; \tag 1$

thus, for any $r \in R$,

$r = r1 = r((1 - e) + e)r = r(1 - e) + re. \tag 2$

If now

$r \in A(e), \tag 3$

we have

$re = 0, \tag 4$

and thus via (2),

$r = r(1 - e) \in (1 - e), \tag 5$

the principal ideal generated by $1 - e$. We thus have

$A(e) \subseteq (1 - e); \tag 6$

likewise, for

$r \in (1 - e), \tag 7$

we have

$r = s(1 - e) \tag 8$

for some $s \in R$; then

$re = s(1 - e)e = s(e - e^2) = s \cdot 0 = 0, \tag 9$

whence

$r \in A(e), \tag{10}$

showing

$(1 - e) \subseteq A(e); \tag{11}$

thus we have

$A(e) = (1 - e) \tag{12}$

by virtue of (6) and (11); $A(e)$ is the principal ideal generated by $1 - e$.