Let $R$ be a commutative ring. Let $M_1,M_2,M_3$ be indecomposable $R$ - modules such that $l(M_1),l(M_2), l(M_3) \geq s$, where $l$ denotes the length of a module. Let $f_1: M_1 \rightarrow M_2$, $f_2: M_2 \rightarrow M_3$ be either injective oder surjective module homomorphisms.

Then $l($im$(f_2f_1))\geq s$.

My ideas:

  1. Since $f_1,f_2$ are either injective or surjective (but not bijective), it follows from the additivity from $l$, that it must hold $l(M_1) \neq l(M_2)$ and $l(M_2) \neq l(M_3)$.

  2. If for $i \in \{ 1,2\}$ $f_i$ is injective $\Rightarrow l($im$(f_i))=l(M_i) \geq s$. If for $i \in \{ 1,2\}$ $f_i$ is surjective $\Rightarrow l($im$(f_i))=l(M_{i+1}) \geq s$

  3. Assuming that $f_1$ is surjective: im$(f_2f_1)=f_2(f_1(M_1))=f_2(M_2)=$im$(f_2)$ and therefore im$(f_2f_1) \geq s$

I don't know how to proceed in the case that $f_1$ is injective. Can someone help?


If you allow $f_1$ to be injective and $f_2$ to be surjective, then the result is false. For example, take $R=k[t]/(t^2)$ with $k$ a field, $M_1=k=M_3$ and $M_2=R$. Then $s=1$ and we have a short exact sequence $$ 0 \to k \to R \to k \to 0 $$ so the composition $f_2f_1:k\to k$ is zero.

If both are injective, then $s\leq l(M_1)\leq l(f_2f_1(M_1))$, and if both are surjective, then $l(f_2f_1(M_1))\geq l(M_3)\geq s$.

  • $\begingroup$ Thank you! Just a quick question: is it possible to show that there exist module homomorphisms $f_1:M_1 \rightarrow M_2, f_2:M_2 \rightarrow M_3$ such that im$(f_2f_1)\geq s$? $\endgroup$ – mathStudent Nov 4 '20 at 14:39

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