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Let $A$ be a ring and let $P$ be a finitely generated $A$-module.

Show that the following conditions are equivalent:

i)$\exists \ A$-module $Q$ and $r\in \mathbb{N}$ and $P\oplus Q \cong A^{\oplus r}$

ii) For any $A$- module $M$ and any surjective $A$-module homomorphism $\pi:M\rightarrow P$, there exist an $A$-module homomorphism $s : P \rightarrow M$ such that $\pi \circ s = id_P$

iii) For any $A$-Module $M, N$ and any $A$-module homomorphisms

$f: P\rightarrow N$ and $g : M\rightarrow N$ with $g$ surjective

There exist an $A$-module homomorphism $h:P\rightarrow M$ such that $g \circ h = f$

I'm facing difficulties in showing (i) $\rightarrow$ (ii).

Any help/insights will be deeply appreciated

PIandpie
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  • There are several questions whose solutions should help you: http://math.stackexchange.com/q/1155248/29335 is probably the one to start with. – rschwieb Mar 17 '17 at 11:07

1 Answers1

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If you want to prove (ii) and you are given (i), consider the surjection $$M \oplus Q \to P \oplus Q = A^{\oplus r}.$$ If this map is split by $s=(s_1,s_2)$, then the restriction of $s_1$ to the first summand is the desired splitting of $M \to P$.

Hence you reduced to the case $P=A^{\oplus r}$. This case is easy, because you can take any basis element of $A^{\oplus r}$ and map it to some pre-image.

MooS
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  • I know I can show that $M \cong P\oplus ker(\pi)$, but how does this reduce to the case you specify ? – PIandpie Mar 17 '17 at 08:24
  • If you have already shown $M \cong P \oplus \operatorname{ker}(\pi)$, you are done anyway, because the inclusion $P \to P \oplus \operatorname{ker}(\pi)$ is the desired map $s$. – MooS Mar 17 '17 at 08:26