$\DeclareMathOperator\Coker{Coker}$Assume the exact sequence $$A \xrightarrow{f} B \xrightarrow{} \Coker f \xrightarrow{} \{0\} $$ where $A$ is a finitely generated module and $B$ is a finitely presented module. Is it true that $\Coker f$ is finitely presented module ? In general, is the quotient of a finitely presented module to a finitely generated submodule, finitely presented ?

  • $\begingroup$ What precisely are the categorical definitions of finitely generated, exact sequence, finitely presented? You might also throw in the definition of projective and the knowledge that free modules are projective. $\endgroup$ – Barbara Osofsky Feb 20 '13 at 3:42
  • $\begingroup$ This is easy to see if $B$ is projective. I have not even been able to find the answer online for the general case though. $\endgroup$ – Alex Youcis Feb 20 '13 at 5:01

This is true.

I'll denote the ring by $R$, while remaining silent about whether these are left or right modules.

Write $B = R^n/I$ for some finitely generated submodule $I \subseteq R^n$. Then the image of $f \colon A \to B$ is of the form $J/I$ for some submodule $J \subseteq R^n$, and $J/I$ is a finitely generated module. Then $J$ is finitely generated as well: taking finitely many elements of $J$ whose images generate $J/I$, along with a finite generating set for $I$, will give a finite generating set for $J$.

Thus $Coker(f) \cong R^n/J$ is finitely presented.

If you're looking for a reference, see Exercise 4.8 in T.Y. Lam's book Exercises in Modules and Rings.

  • 1
    $\begingroup$ The same proof shows: If $A \to B \to C \to 0$ is exact, $A$ is of finite type and $B$ is of finite presentation, then $C$ is of finite presentation. $\endgroup$ – Martin Brandenburg Sep 17 '13 at 9:55
  • $\begingroup$ @MartinBrandenburg, indeed! $\endgroup$ – Manny Reyes Sep 17 '13 at 13:13

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