# quotient of finitely presented module

$\DeclareMathOperator\Coker{Coker}$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 ?

• 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. Commented Feb 20, 2013 at 3:42
• 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. Commented Feb 20, 2013 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.

• 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. Commented Sep 17, 2013 at 9:55
• @MartinBrandenburg, indeed! Commented Sep 17, 2013 at 13:13

By A→B→Coker f →0, we get an exact sequence 0→k→B→Coker f →0, where k is the image of f, so k is a finitely generated module.Take a projective resolution ...→P1→P0→k→0，where P0 finitely generated.Also take a projective resolution ...→Q1→Q0→B→0，where Q0,Q1 finitely generated.Then the mapping cone ...→ P0⊕Q1→Q0→Coker f→0 is a projective resolution of Coker f, where Q0, P0⊕Q1 are finitely generated projective module, so Coker f is finitely presented. For the mapping cone, you can learn from triangulated category or homological algebra or in Wikipedia https://en.wikipedia.org/wiki/Mapping_cone_(homological_algebra). I hope the upper will help you.