If $L\rightarrow M\rightarrow N$ is a short exact sequence of A-modules and $L$ and $N$ are finite over A, then so is M.
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2are you trying to prove this? If so, what have you tried? – MITjanitor Mar 09 '13 at 20:32
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here I don't know how to make $M$ into a quotient of a f.g. A-module. – Alex Mar 09 '13 at 20:33
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Calc1DropOut: yes. – Alex Mar 09 '13 at 20:34
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1And what about his second question? – Julian Kuelshammer Mar 09 '13 at 20:36
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1This is an abstract duplicate of this question (in the question above, just replace $L$ and $N$ with the appropriate pieces making this a short exact sequence). – Zev Chonoles Mar 09 '13 at 20:39
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Hint: Take generating sets for $L$ and $N$. Push the generating set of $L$ into $M$ and for each element of the generating set in $N$ choose a preimage in $M$. These two sets together are a finite set in $M$ and you should be able to show that it generates.
Jim
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