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Let $0 \rightarrow A \xrightarrow{\psi} B \xrightarrow{\sigma} C \rightarrow 0$ be a short exact sequence of $R$-modules.

If there is a homomorphism $\rho :C \longrightarrow B$ such that $\sigma \circ \rho$ is the identity in $C$, how do you prove that $B=\psi(A)\bigoplus \rho(B)$?

It is straightforward to check that $\psi(A)\cap \rho(B)=0$, I don't know how to show that $\psi(A) + \rho(B) = B$ ?

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Going back to the group case, are you familiar with the fact that a short exact sequence with a section $\iff$ semidirect product? –  mixedmath Mar 19 '12 at 20:28

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up vote 3 down vote accepted

Hint. $b - \rho(\sigma(b))\in\mathrm{Ker}(\sigma) = \mathrm{Im}(\psi)$.

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thanks! One more question, if $f \frac{\mathbb{Z}}{n\mathbb{Z}} \longrightarrow \mathbb{Z}$ is a $\mathbb{Z}$-module homomorphism,then why $f(1) = 0$? Thanks. –  Jr. Mar 19 '12 at 20:52
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If $n\neq 0$, Because $1+n\mathbb{Z}$ is a torsion element of the domain; and torsion elements map to torsion elements. What are the torsion elements of $\mathbb{Z}$? (If $n=0$, then the statement is false). –  Arturo Magidin Mar 19 '12 at 20:53
    
surely, "torsion elements map to torsion elements" because they have the same order?or it's because another reason? –  Jr. Mar 19 '12 at 21:00
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No, they don't have to have the same order (think $(\mathbb{Z}/4\mathbb{Z})\to(\mathbb{Z}/2\mathbb{Z})$. But certainly, the image of a torsion element has nontrivial annihilator. –  Arturo Magidin Mar 19 '12 at 21:04
    
oh I see it now.Thanks. –  Jr. Mar 19 '12 at 21:12

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