Short exact sequence with four terms

Question

We want to build an exact sequence $$ 0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow D \longrightarrow 0 $$ such that $A \oplus C \cong B \oplus D$ and other in which that property does not hold.

Is it possible to obtain this with all modules not being null? I am trying to extend the idea of inclusion and projection with the direct sum in the middle term but doesn't seem to work.

julio_es_sui_glace · Accepted Answer · 2023-06-26T00:41:15.470

1

$R$ a ring, take $A = R$, $B = R \oplus R$, $C = R \oplus R$ and $D = R$. We name our morphism in order $\alpha$, $\beta$, $\gamma$. Take $$\alpha = 1_R \oplus 0$$ $$\beta \circ(1_R\oplus 0) = 0 \oplus 0 , \, \beta \circ(0\oplus 1_R) = 1_R \oplus 0$$ $$\gamma \circ (1_R \oplus 0) = 0 , \, \gamma\circ(0 \oplus 1_R) = 1_R$$

For the non-split version, take $A = \mathbb{Z}$, $B = \mathbb{Z} \oplus \mathbb{Z}$, $C = \mathbb{Z}$ and $D = \mathbb{Z}/2\mathbb{Z}$ And $$\alpha(1) = 1\oplus 0$$ $$\beta(1\oplus 0) = 0 , \, \beta(0\oplus 1) = 2 $$ $$\gamma(1) = 1$$

You can easily extend the length of your sequence for both examples

edited Jun 26 '23 at 00:41

answered Jun 25 '23 at 18:05

julio_es_sui_glace

3,279

Yes, thanks. I think it should be $\alpha(n) = (n, 0)$ in both cases instead of $\gamma$. Also, nice remark! – Paolo Jove Jun 26 '23 at 00:28
Yes a small typo :) – julio_es_sui_glace Jun 26 '23 at 00:32
And $\gamma(n, m) = m$ in the first example :) – Paolo Jove Jun 26 '23 at 00:35
Yes, also I used $1_R$ as the identity function on $R$ to avoid imposing $R$ unitary :) – julio_es_sui_glace Jun 26 '23 at 00:42
Beware: you can’t give general counterexamples since for some rings like fields, you only have split sequences! – julio_es_sui_glace Jun 26 '23 at 00:45

Short exact sequence with four terms

1 Answers1