Please, give an example of a module $M$ such that $M$ is primeless (i.e. without prime submodule) and projective. Thanks for your attention.
An example of a primeless (i.e. module without prime submodule) and projective module
1
$\begingroup$
commutative-algebra
1 Answers
1
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.
-
1@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. – 2012-05-27
-
0Dear @Michael, you did well, of course, but how could m. sam. with a reputation of only 23 edit this answer ? – 2012-08-12
-
0@GeorgesElencwajg : He did submit an edit, which I rejected. – 2012-08-12
-
0Ah, now I understand. Thanks for the explanation, @Michael. – 2012-08-12