Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Please, give an example of a module $M$ such that $M$ is primeless (i.e. without prime submodule) and projective. Thanks for your attention.

share|cite|improve this question

I learn the notion of "prime module" from wiki's article associated prime. According to wiki, it only needs to give a projective module which has no associated prime ideal.

Let $A=k[x_1,x_2,\ldots]/(x_i^2\mid i=1,2,\ldots)$ which is a quotient ring of the infinitely many indeterminates polynomial ring over a field $k$. Then $A$ has only one prime ideal $\mathfrak{m}=(x_1,x_2,\ldots)$. It is not an associated prime, you cannot find an $f\in A$, such that $(0:f)=\mathfrak{m}$. And $A$ is a free module as $A$-module.

share|cite|improve this answer
@m. sam. : You wrote: "Thanks for your answer, but I am looking for a module $M$ such that $M$ is not a free or a multiplication module and moreover $M$ is primeless and projective." The proper place for such a comment is down here in the comments section. I've deleted it from the answer. – Michael Hardy May 27 '12 at 18:09
Dear @Michael, you did well, of course, but how could m. sam. with a reputation of only 23 edit this answer ? – Georges Elencwajg Aug 12 '12 at 8:34
@GeorgesElencwajg : He did submit an edit, which I rejected. – Michael Hardy Aug 12 '12 at 15:27
Ah, now I understand. Thanks for the explanation, @Michael. – Georges Elencwajg Aug 12 '12 at 18:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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