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 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!
 A: If $\langle r\rangle=Rr$ is a direct summand of $R$, then we have a split exact sequence
$$\{0\}\to Rr\to R\to R/Rr\to\{0\}$$
hence there exists an $R$-module homomorphism $\pi:R\to Rr$ such that $\pi(ar)=ar$ for all $a\in R$.
In particular, $\pi(r)=r$, hence $\pi(r^n)=r^{n-1}\pi(r)=r^n$ for all $n>0$.
Let $\pi(1)=ur$ for some $u\in R$.
Let $n$ be the smallest $n>0$ such that $r^n=0$.
Then $r^{n-1}=\pi(r^{n-1})=r^{n-1}\pi(1)=ur^n=0$ - a contradiction.
A: 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).
A: Commutative case: 
If $(r)$ were a direct summand of $R$, it would be generated by an idempotent $e$, that is, $eR=(r)$.
This is an elementary fact about summands of rings with identity. The idea is that if $A\oplus B=R$, then $e+f=1$ for some unique $e\in A$, $f\in B$ and that $e,f$ are idempotent generators of $A,B$ respectively.
But everything in $(r)$ is nilpotent, including $e$. The only nilpotent idempotent is $0$, so that would imply $r=0$.

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.

Well, first and foremost, the thing you said earlier I know that $(r)$ is then a nilpotent ideal. is completely false in the general noncommutative case.
For example, if $R=M_2(\mathbb Q)$ and $r=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, then $(r)=R$ is in fact a summand.
Even if you switch to one-sided ideals, it is still not true that $rR$ is nilpotent: $rR=\begin{bmatrix}1&0\\0&0\end{bmatrix}R$ and $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is not nilpotent, and the right ideal is still a summand.
A: This might not be what you're looking for, but if $(r)$ is a direct summand of $R$, then $(r)$ is projective, so it is torsion free. But $(r)$ clearly has torsion since it is nilpotent.
