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

Michael Hardy
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m. sam.
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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.

Michael Hardy
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wxu
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    @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 08:34
  • @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