Let $R$ be a commutative ring (with $1$) such that every non-zero divisor in $R$ is a unit (see Rings in which every non-unit is a zero divisor for various stabs at what these are called). Let $M$ be a finitely generated $R$-module.

My question:

Can we conclude that every non-zero divisor in $\mathrm{End}(M)$ is a unit?

For example: When $M$ is a free module, we have $\mathrm{End}(M) \cong R^{n \times n}$ is a matrix algebra and so every non-zero divisor is a unit (see Do these matrix rings have non-zero elements that are neither units nor zero divisors?).

A few notes:

  1. Even matrix algebras can misbehave when $R$ is non-commutative. Also, if $R$ has a non-zero divisor which is not a unit, then $\mathrm{End}(R)=R^{1 \times 1}=R$ has a non-zero divisor which is not a unit (thus the assumptions on $R$).
  2. If $V$ is an infinite dimensional vector space (over some field). Then $\mathrm{End}(V)$ has non-zero divisors which aren't units (thus the "finitely generated" assumption).
  3. If the answer to the question is "No" (I'd like a counter-example), then I would like to know under what circumstances $\mathrm{End}(M)$ does have this property. What assumptions on the module will force this to hold? (e.g. This holds when $M$ is a ??? module -- like flat or projective or something.) Or what assumptions on the ring will force this to hold for all modules? (e.g. This obviously holds when $R$ is a field.)
  4. Motivation and background: I've been working with a few undergraduates on a related problem. We tripped over this question and I have no idea whether this is true or not (I'm not a ring theorist and have a feeling this may be a difficult question to entirely resolve).

Thanks for reading my question! We (my group of undergrads and myself) appreciate your help!

  • 1
    $\begingroup$ Inspired by the answer, which is the canonical example of a non-Cohen-Macauley ring: If we demand the ring be Cohen-Macauley, then it has Krull dimension equal to its depth, which is $0$, so it is Artinian, so its modules are Artinian and clearly have this property. But presumably you don't want to just consider Artinian rings? $\endgroup$
    – Will Sawin
    Mar 11, 2013 at 1:56
  • 2
    $\begingroup$ Note that there are classes of modules $M$ for which the statement holds regardless of $R$ having this property or not: 1) $M$ indecomposable of finite length (Fitting's lemma); 2) $M$ semisimple, which makes $End_R(M)$ von Neumann regular; 3) $M$ nonsingular and continuous ($\Leftarrow$ quasi-injective $\Leftarrow$ injective), where again $End_R(M)$ is von Neumann regular (and self-injective). $\endgroup$ Mar 11, 2013 at 19:28

2 Answers 2


Here is a small positive result: If $R$ is Artinian, $End_R(M)$ is an $R$-algebra which is f.g. as $R$-module, hence Artinian, hence classical (as I call these rings, following Lam).

On the other hand, as soon as the Krull dimension of $R$ is $\geq 1$, $M = R/p$ for a non-maximal prime ideal $p$ will give a counterexample, as in Graham's answer.

Added: I think if $R$ is Noetherian (hence semilocal by a theorem of Faith, see Lam's Lectures on Modules and Rings theorem 8.31 p. 283), the assertion is true for $M$ f.g. projective (= f.g. flat).
Because then $R \simeq \prod_{i=1}^{n} R_i$ with indecomposable semilocal $R_i$ with corresponding idempotents $e_i$, $End_R(M) \simeq \prod End_{R_i}(e_i M)$ and $e_iM$ is f.g. projective over $R_i$. We can thus reduce to the indecomposable case, and here f.g. projectives are free (this remains true without "f.g.", says Hinohara), so we are done by the OP's second link.

  • $\begingroup$ These modules are neither flat nor projective. Could a criterion like those mentioned in the question solve the problem? $\endgroup$
    – Will Sawin
    Mar 11, 2013 at 2:18
  • $\begingroup$ @Will, see the addition for the Noetherian case. I think there should also be something to say for $R$ non-Noetherian and zerodimensional, or at least von Neumann regular, but I don't see it right now. For general non-Noetherian rings, I have no idea. $\endgroup$ Mar 11, 2013 at 19:48
  • $\begingroup$ Torsten, thanks for your response. I've been meaning to read through Lam's text and just haven't gotten around to it yet. Thanks agian for your answer, you've given my group plenty to dig through! :) $\endgroup$
    – Bill Cook
    Mar 12, 2013 at 1:37

Counterexample: $R=k[[x,y]]/(x^2,xy)$ and $M = R/(x) \cong k[[y]]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.