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

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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).