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An $R$-module $M$ is said to be weakly Laskerian if the set of associated primes of any quotient module of M is finite.

Let $M$ be a weakly Laskerian $R$-module. Since $M\cong M/0$ then $Ass(M)$ is finite. Is this right?

Thanks.

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6 
 Do you really have any doubt? – Martin Brandenburg Aug 6 2011 at 16:02
1 
 Because i think, in the definition of weakly Laskerian module, if $M/N$ is quotient module of $M$, that mean $N$ is a proper submodule of $M$. Thanks. – minhtringuyen Aug 7 2011 at 15:49
Fortunately, zero is a proper submodule of $M$. I think math.stackexchange.com is a more appropriate home for this question. – S. Carnahan♦ Aug 8 2011 at 3:54

closed as too localized by Andres Caicedo, Qiaochu Yuan, Andreas Blass, Simon Thomas, S. Carnahan♦ Aug 8 2011 at 3:52

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