Let $R$ be a commutative ring with $1$ and $M$ is an $R$-module.

Let $I$ be an ideal of $R$ contained in $\mathrm{Ann}(M)$.

Consider the $R/I$-module $M$.

I know the fact that the $R/I$-module $M$ is noetherian (artinian) iff the $R$-module $M$ is noetherian (artinian) because of the property that is submodule is still submodule when we change rings.

Here is my question:

Is that true that the $R$-module $M$ has property $P$ iff the $R/I$-module $M$ has property $P$?

From my intuition, I would say that is true but can't find an explicit proof. I looking for the proof or a counterexample.

Thanks in advance.

  • 1
    $\begingroup$ Certainly not true. $R/I$ is a free module when regarded as a module over the rign $R/I$, but not when regarded as a module over the ring $R$ unless $I=(0)$. $\endgroup$ – mathguy Sep 25 '16 at 19:07
  • $\begingroup$ thank you ! do you find another counterexample ? $\endgroup$ – Anh_Rose 1210 Sep 25 '16 at 19:22
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    $\begingroup$ Plenty. Flatness. Injectivity (being an injective module). Reflexivity. Whenever a property IS preserved between $R$ and $R/I$ that is a reason for celebration, it's a well-known theorem with a well-known name. $\endgroup$ – mathguy Sep 25 '16 at 19:52
  • $\begingroup$ can you give me the name of this theorem ?, so I can learn more about it. $\endgroup$ – Anh_Rose 1210 Sep 25 '16 at 20:57
  • $\begingroup$ I didn't mean "one" theorem. I meant, whenever a property (ANY property) is preserved between $R$ and $R/I$ module structure, that's an important result. Why do you care about "abstract" (random, "whatever") properties that are or are not preserved? You should only worry about properties you are interested in for some reason. $\endgroup$ – mathguy Sep 25 '16 at 23:25

In general no. For a counterexample, consider an $R$-module $M$, such that $\mathrm{Ann}_R(M)\neq\{0\}$, that is a non-faithful $R$-module. Then, if you take $I=\mathrm{Ann}_R(M)$ you will get a faithful $R/I$-module $M$, i.e., one for which $\mathrm{Ann}_{R/I}(M)=\{0\}$.


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