# In a short exact sequence of modules, the extremes are finitely presented. Why is the central module finitely presented?

I am currently stuck in the following problem, which is an excercise of a lecture of module theory:

Problem. Let $R$ be a unital ring and $$(*)\quad 0\rightarrow M\overset{f}{\rightarrow}N\overset{g}{\rightarrow}P\rightarrow 0$$ be a short exact sequence of right $R$-modules. If $M$ and $P$ are finitely presented, then $N$ is finitely presented.

All I have so far is that by the exactness of $(*)$ and the following exact sequences \begin{align} &R^m\rightarrow R^n\rightarrow M\rightarrow 0\\ &R^k\rightarrow R^l\rightarrow P\rightarrow 0 \end{align} it can be seen that $\ker g$ and $N/\ker g$ are both finitely presented. But I can hardly figure out what to do next. Does it follow from this fact that $N$ is finitely presented?

Here I guess I worked out another solution. The conditions are equivalent to \begin{align} \require{AMScd} \begin{CD} @. R^m @. \ @. R^l @.\\ @. @VVV @. \ @VVV @.\\ @. R^n @. \ @. R^k @.\\ @. @V{\alpha}VV @. @VV{\gamma}V @.\\ 0@>>> M@>>{f}> N@>>{g}> P@>>> 0\\ @. @VVV @. @VVV @.\\ @. 0 @. @. 0 @. \end{CD} \end{align} Let $\iota\colon R^n\to R^n\oplus R^k$ and $\pi\colon R^n\oplus R^k\to R^k$ be respectively the natural embedding and the natural projection. By $R^k$ is free and thus projective, as the following diagram shows, \begin{align} \require{AMScd} \begin{CD} @. R^k @.\\ @. @VV{\gamma}V @.\\ N @>>{g}> P @>>> 0 \end{CD} \end{align} $\gamma$ can be lifted to a homomorphism $\tilde\gamma\colon R^k\to N$, such that $g\circ\tilde\gamma=\gamma$. Set $\tilde\alpha=f\circ\alpha\colon R^n\to N$. Define a homomorphism $\beta\triangleq \tilde\alpha\oplus\tilde\gamma\colon R^n\oplus R^k\to N$ by $\beta(x, y)=\tilde\alpha(x)+\tilde\gamma(y)$. Then we have \begin{align} \require{AMScd} \begin{CD} @. R^m @. \ @. R^l @.\\ @. @VVV @. \ @VVV @.\\ 0 @>>> R^n @>{\iota}>> R^n\oplus R^k @>{\pi}>> R^k @>>> 0\\ @. @V{\alpha}VV @VV{\beta}V @VV{\gamma}V @.\\ 0@>>> M@>>{f}> N@>>{g}> P@>>> 0\\ @. @VVV @. @VVV @.\\ @. 0 @. @. 0 @. \end{CD} \end{align} This diagram is commutative. Indeed, for each $x\in R^n$, \begin{align} \beta\circ\iota(x)=\beta(x,0)=\tilde\alpha(x)=f\circ\alpha(x); \end{align} for each $(x, y)\in R^n\oplus R^k$, \begin{align} g\circ\beta(x, y)=&g(\tilde\alpha(x)+\tilde\gamma(y))=g(\tilde\alpha(x))+g(\tilde\gamma(y))\\ =&g\circ f\circ\alpha(x)+\gamma(y)=\gamma(y)\\ =&\gamma\circ\pi(x,y). \end{align} Then the snake lemma entails that \begin{align} \ker\alpha\to\ker\beta\to\ker\gamma\to 0\to\mathrm{coker}\beta\to 0 \end{align} is exact. Then $\beta$ is surjective and hence $R^n\oplus R^k\overset{\beta}{\to} N\to 0$ is exact. The same argument applies to \begin{align} \require{AMScd} \begin{CD} @. R^m @>>> R^m\oplus R^l @>>> R^l @>>> 0\\ @. @VVV @VVV @VVV @.\\ 0@>>> \ker\alpha@>{\iota}>> \ker\beta@>{\pi}>> \ker\gamma@>>> 0\\ @. @VVV @. @VVV @.\\ @. 0 @. @. 0 @. \end{CD} \end{align} and we can get that \begin{align} R^m\oplus R^l\to R^n\oplus R^k\to N\to 0 \end{align} is exact, and therefore $N$ is finitely presented.

• Sorry for the late reply, yet I wanna point out that this is the same trick used in proving the Horseshoe lemma in homological algebra, see Rotman's An Introduction to Homological Algebra, Proposition 6.24, for example. May 8, 2022 at 8:06

This is exercise 4.8(2) in Exercises in modules and rings by T.Y. Lam. We can prove it as follows:

First, observe that since $M, P$ are finitely generated, $N$ is also finitely generated: one can show that if $m_1,\ldots,m_n$ generate $M$ and $p_1,\ldots,p_k$ generate $P$, then $f(m_1),\ldots,f(m_n)$, $g^{-1}(p_1),\ldots, g^{-1}(p_k)$ generate $N$.

Therefore $N\cong R^n/X$ for some $n\in\mathbb{N}$ and some $R$-module $X$. The aim now is to prove that $X$ is finitely generated.

Since $M$ is injected in $N$, there is some $Y\leq R^n$ such that $M\cong Y/X$. Therefore $$P\cong N/M\cong (R^n/X)/(Y/X)\cong R^n/Y.$$ This implies that $Y$ is finitely generated: We have the exact sequence $$0\rightarrow Y\rightarrow R^n\rightarrow P\rightarrow 0,$$ but we know that $P$ is finitely presented, so we also have $$0\rightarrow K\rightarrow R^k\rightarrow P\rightarrow 0$$ with $K$ finitely generated. Since both $R^n$ and $R^k$ are projective, by Schanuel's lemma we get $$R^n\oplus K\cong R^k\oplus Y$$ with $K$ finitely generated, so $Y$ is finitely generated as well.

Now we have an exact sequence for $M$ $$0\rightarrow X\rightarrow Y\overset{\alpha}{\rightarrow} M\rightarrow 0$$ in which $Y$ is finitely generated and $M$ finitely presented. This implies that $X$ is finitely generated: Since $Y$ is finitely generated, we can consider the short exact sequence $$0\rightarrow \ker(\alpha\beta)\rightarrow R^k\overset{\alpha\beta}{\rightarrow} M\rightarrow 0$$ where $R^k\overset{\beta}{\rightarrow} Y$. By the same Schanuel's argument as before (since $M$ is finitely presented), this implies $\ker(\alpha\beta)$ is finitely generated, so $\ker(\alpha)\cong X$ is also finitely generated, as $\beta$ is surjective: The map $\ker(\alpha\beta)\rightarrow\ker(\alpha)$ such that $y\mapsto\beta(y)$ is well defined ($\beta(y)\in\ker(\alpha\beta)$ because $\alpha\beta(y)=0$ as $y\in\ker(\alpha\beta)$) and is an epimorphism (if $x\in\ker(\alpha)$ then there exists $y\in R^k$ such that $\beta(y)=x$; then $\alpha\beta(y)=0$ and thus $y\in\ker(\alpha\beta)$). Lastly, an epimorphic image of a finitely generated module is finitely generated.

• Thank you! I guess I have understood your nice proof but for one point, that how that $\ker(\alpha)$ is finitely generated can be obtained from the surjectiveness of $\beta$? I only know it implies that $\ker(\alpha)=\beta(\ker(\alpha\beta))$...so could you please kindly explain it a little? Nov 7, 2017 at 15:04
• @josephz Yes, that merits some explanation! I have edited the answer to add it Nov 7, 2017 at 16:13
• Oh I see...Now I have understood the proof~ Thank you again! Nov 7, 2017 at 17:06