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Let $R$ be a ring and $M$ an $R$-module. Suppose I know that $$ M/A \cong M / B $$ as $R$-modules and $A,B$ are submodules of $M$ such that $A \subseteq B$.
Is it then the case that $A = B$?

I was trying to figure out details of the accepted solution of Proving that the tensor product is right exact

We have
$$A\otimes_R M\xrightarrow{\alpha\otimes Id}B\otimes_R M\xrightarrow{\beta\otimes Id}C\otimes_R M\rightarrow 0$$ and from the solution it follows that: let $D$ be the image of $\alpha \otimes \operatorname{Id}$ then $(B \otimes M)/D \to C \otimes M$ is an isomorphism (assuming we have already shown $\beta \otimes \operatorname{Id}$ is surjective). It follows that $(B \otimes M)/D \cong (B \otimes M)/\ker (\beta \otimes \operatorname{Id})$, but I was wondering how I can show that $D = \ker (\beta \otimes \operatorname{Id})$ in fact equal assuming $D \subseteq \ker (\beta \otimes \operatorname{Id})$.... thank you

Johnny T.
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    The statement in the title is not true for infinite dimensional vector spaces. Pick $M$ any infinite dimensional vector space, $A={0}$ and $B$ any nontrivial finite dimensional subspace of $M$. Then $A$ and $B$ are not even isomorphic. – Severin Schraven Sep 06 '20 at 20:13
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    Regarding the right exactness of the tensor product: the kernel of the map $(B\otimes M)/D \to C\otimes M$ is $\ker(\beta\otimes \operatorname{Id})/D$. So if the map is an isomorphism (in particular injective), then $\ker(\beta\otimes \operatorname{Id})/D = 0$. But this means $\ker(\beta\otimes\operatorname{Id}) = D$. – Claudius Sep 06 '20 at 20:55
  • @Claudius It makes sense now, thank you for the clarification! – Johnny T. Sep 07 '20 at 21:33
  • This is late, but a warning: the question should have been asked in the context of abelian groups, not modules, as $M \otimes N$ is in general not a module. – John Cavanaugh Sep 29 '24 at 04:44

2 Answers2

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The answer is "no". Take the free module $M$ freely generated by $x_i$ for positive integers $i$ and its factor-module $M'$ freely generated by $x_{2n}$,. the homomorphism consists of killing all $x$'s with odd indices. Then $M, M'$ are isomorphic.

Bernard
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markvs
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The answer is yes if what you are assuming is an isomoprhism is the canonical map $M/A\longrightarrow M/B$ induced by the inclusion $A\subseteq B$: the kernel of this map if $B/A$, which is zero if and only if $A=B$. This is the situation in your question.

Pedro
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