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Let $0 \to M_1 \to M \to M_2 \to 0$ be an exact sequence of $A$-modules.

i) Prove: If $M_1$ and $M_2$ are finitely generated, then $M$ is too.

ii) Let $M$ and $N$ be sub-modules of an $A$-module $L$. Assume that $M+N$ and $M\cap N$ are finitely generated $A$-modules. Prove that $M$ and $N$ are finitely generated $A$-modules.

If have proved the first part, but i don't see how to do the second part. I tried to use the first part, so to find an exact sequence with $M$ in the middle and finitely generated modules in place of $M_1$ and $M_2$, but I can't find a surjective map $M \to P$ (for some f.g. $A$-module) in the sequence:

$$0 \to M\cap N \to M \to P \to 0$$

any hints/help would be appreciated.

user26857
  • 53,190
Algebra
  • 11

2 Answers2

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Simply take $$P= M/(M \cap N)$$ and recall that this is finitely generated since there is a canonical isomorphism $$M/(M \cap N) \cong (M+N)/N$$

Crostul
  • 37,500
0

For i) note that the sum of $A^n \twoheadrightarrow M_1 \to M$ and a lift of $A^m \twoheadrightarrow M_2$ to $M$, surjects onto $M$.

For ii) note that you get short exact sequences $0 \to M\cap N \to M \to (M+N)/N\to 0$.

Daniel Valenzuela
  • 6,405