TITLE: If $r$ is a nilpotent element of a commutative ring $R$ then $(r)$ is not a direct summand of $R$ as an $R$-module.
QUESTION [6 upvotes]: I am having trouble proving that if $r$ is a nilpotent element of a commutative ring $R$ then $(r)$ is not a direct summand of $R$ as an $R$-module. I know that $(r)$ is then a nilpotent ideal. I understand that it doesn't work when $R$ is not commutative, but only by an example - I don't see where it actually breaks down. 
Thank you in advance for any help!

REPLY [2 votes]: I'm a tad rusty on projective modules and such, so I worked up a proof based on "first principles", only using basic definitions and so forth:
I assume $r \ne 0$; in the case $r = 0$, it is trivial that $R = (0) \oplus R$.
Suppose that $(r)$ were a direct $R$-modular summand of $R$; that is,
$R = (r) \oplus M; \tag 1$
then by definition
$R = (r) + M, \; (r) \cap M = \{0\}, \tag 2$
where $M \subset R$ is also an $R$-modular summand; $M$ being an $R$-submodule of $R$ requires
$RM \subset M; \tag 3$
that is, $M$ itself is an ideal of $R$; furthermore, $M$ must be proper in $R$, that is
$M \subsetneq R, \tag 4$
since $r \ne 0$ forces, via (2),
$r \notin M. \tag 5$
Now. again by virtue of (2), $1 \in R$ must be of the form
$1 = ar + m, \tag 6$
for some
$a \in R, \; m \in M; \tag 7$
thus,
$1 - ar = m \in M; \tag 8$
but the hypothesis $r^n = 0$ implies $m = 1 - ar$ is invertible in $R$, since we have the identity
$m \displaystyle \sum_0^n (ar)^k = (1 - ar) \sum_0^n (ar)^k = \sum_0^n (ar)^k - \sum_1^{n + 1} (ar)^k$
$= 1 - (ar)^{n + 1} = 1 - a^{n + 1}r^{n + 1} = 1; \tag 9$
and now we further infer from (8) that
$1 = m \displaystyle \sum_0^n (ar)^k \in M \Longrightarrow M = R, \tag{10}$
since
$\forall b\in R, \; b = b(1) \in M; \tag{11}$
but $M = R$ contradicts (4), so we cannot have (1).