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Problem

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Corollary 3.13

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Partial Proof

If $M'$ and $M$ are finitely presented (and therefore finitely generated) it's obvious that $M''$ is finitely presented. If $M$ is finitely presented, then $M'$ is isomorphic to a submodule of $M$, which is finitely presented.

Now suppose $M'$ and $M''$ are finitely presented. I need to show that $M$ is finitely presented...?

Because each module is finitely presented, it follows by definition that they have Euler characteristics.

Questions

  1. How is Corollary 3.13 relevant when one of these modules are finitely presented?
  2. Generally not sure how to tackle this problem, though I did get (i) and (ii).
IsaacR24
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1 Answers1

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Given the short exact sequence $0 \to M' \to M \to M'' \to 0$, we can choose free resolutions of each so that they fit into short exact sequences at each level: $0 \to F_m' \to F_m \to F_m'' \to 0$. Let's assume that the Euler characteristic exists for two of the three modules. Then there is an $n$ so that two of $F_n'$, $F_n$, and $F_n''$ are zero, and therefore the third is also zero.

It remains to check that for each $m$, if two of $F_m'$, $F_m$, and $F_m''$ are finitely generated, so is the third.

  • if $F_m'$ and $F_m$ are finitely generated, then $F_m''$ is a quotient of $F_m$ and so it is finitely generated.
  • if $F_m'$ and $F_m''$ are finitely generated, well, we can actually choose free resolutions so that $F_m = F_m' \oplus F_m''$ for each $m$, in which case it is clear that $F_m$ is finitely generated.
  • if $F_m$ and $F_m''$ are finitely generated, then we apply Corollary 3.13 to the short exact sequence $0 \to F_m' \to F_m \to F_m'' \to 0$ to conclude that $F_m'$ is finitely generated.
John Palmieri
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  • I'm not familiar with the result that says "we can choose free resolutions of each so that they fit into short exact sequences at each level". I don't see this result in Rotman's book. Can you elaborate? Otherwise everything makes sense, except just noting that Rotman doesn't address free resolutions until a later chapter. – IsaacR24 Nov 15 '23 at 13:27
  • A free resolution is just an exact sequence as in (ii), except it doesn't have to have only finitely many terms, and each term need not be finitely generated. – John Palmieri Nov 15 '23 at 17:33
  • I don't have access to Rotman's book, but the "Horseshoe Lemma" (https://en.wikipedia.org/wiki/Horseshoe_lemma) is one tool that can help; it deals with the second bulleted item in the list. – John Palmieri Nov 15 '23 at 18:34
  • I see now that the Horseshoe Lemma is the missing piece, and this makes sense to me! For Bullet Point 2, I just note that for any exact sequence $ 0 \to A \to B \to P \to 0$ with $P$ projective that sequence always splits. So your bullet point 2 is true for any choice of free resolution so long as they fit into short exact sequences as you've claimed. – IsaacR24 Nov 16 '23 at 13:19