Let $\mathfrak m$ be a maximal ideal of $R$ and $M$ an $R$-module such that $\mathfrak mM\ne M$. Is it true that $\mathfrak mM$ is a maximal submodule of $M$? Thank you.

(I can see this happen in case $M$ is cyclic.)


Of course not! Take $M=R^2$. Then $M/\mathfrak mM$ is an $R/\mathfrak m$-vectorspace of dimension $2$, so there is a proper subspace $N/\mathfrak mM$ of $M/\mathfrak mM$. This shows that $\mathfrak mM\subset N\subset M$.

  • $\begingroup$ Yes, $M=\left\langle(1,0)+mM,(0,1)+mM \right\rangle $. Thank you! $\endgroup$ – Q.TL Mar 1 '13 at 12:26
  • $\begingroup$ @ YACP. In case $M$ is radical (i.e. $M$ has no maximal submodule), does we have $mM=M$? (m is maximal ideal of $R$) $\endgroup$ – Q.TL Mar 1 '13 at 12:35
  • $\begingroup$ Isn't it even more instructive to consider the case of $R$ being a field and M be an vector space of dimension greater 1. I know it is a special case of your example but it's simpler. $\endgroup$ – Curufin Mar 1 '13 at 14:18
  • $\begingroup$ @user63895 Have you seen this topic? $\endgroup$ – user26857 Mar 1 '13 at 16:08
  • $\begingroup$ @ YACP. Thanks you. $\endgroup$ – Q.TL Mar 2 '13 at 2:55

For $\mathfrak{m}=0$ the question is: Is $\{0\}$ a maximal subspace in every nontrivial vector space? And this is only true in the 1-dimensional case.


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