# length of composition of module homomorphisms

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 $$l(f_2f_1(M_1))\geq l(M_3)\geq s$$.
• 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$(f_2f_1)\geq s$? – mathStudent Nov 4 '20 at 14:39