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I want help to the following problem(especially I want to tell me which is the basic idea behind the solution):

Problem: We have the following assumptions:

  • Let $B$ is projective $R-$module
  • $A$ is $R-$submodule of $B$ and
  • $B/A$ is projective $R-$module.

Show that: $A$ is projective $R-$module.

Solution(my attempt): Let $\psi : C \rightarrow D$ is $R-$module epimorphism and $f : A \rightarrow D$ is $R-$module homomorphism.

Then, we want to find a unique $R-$module homomorphism $g : A \rightarrow C$ such that $\psi \circ g = f$.

Consider the $R-$module epimorphism $i : B \rightarrow A$, (since $B$ is projective $R-$module), then there is unique $R-$module homomorphism $h : B \rightarrow C$ such that $\psi \circ h = f \circ i$.

We have also the canonical $R-$module $\pi : B \rightarrow B/A$ with $\ker\pi = A$.

(This is my attempt)

How can we use that $B/A$ is projective $R-$module?

and after

How can we find this unique $R-$module homomorphism $g : B \rightarrow C$ with $\psi \circ g = f$?

Can you tell me how can I think similar problems, what is the key-idea behind the solution?

Thank you a lot!

TrItOs
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  • Use the following hint: As $B/A$ is projective, the canonical surjection $B \to B/A$ splits. – Geoffrey Trang Dec 21 '23 at 18:08
  • @GeoffreyTrang : Can you explain the "splits". Can you write more details? thank you. – TrItOs Dec 21 '23 at 18:17
  • Splitting $\pi:B \to B/A$ means that there is some $h:B/A \to B$ with $\pi \circ h=\mathrm{id}_{B/A}$ (the identity map). Is that clear now? It is known that the existence of such an $h$ is equivalent to $A$ being a direct summand of $B$. – Geoffrey Trang Dec 21 '23 at 18:24
  • @GeoffreyTrang : Yes, thank you, now it is clear! But I do not understand that "A is the direct summand of B" – TrItOs Dec 21 '23 at 18:35
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    Well, apparently, while I already told you what "splitting $\pi$" means, you still do not know what "direct summand" means either. OK, I'm going to give you the meaning of that as well. It means that there is some submodule $A'$ of $B$ for which $A \oplus A'=B$, i.e., $A+A'=B$ and $A \cap A'=0$. Seriously, do you really need me to then give the meaning of the sum and intersection of submodules as well? – Geoffrey Trang Dec 21 '23 at 19:18
  • @GeoffreyTrang : Thank you a lot , it is very useful! So, we can apply that $B/A$ is projective, then there is unique $R-module$ homomorphism $k : B/A \rightarrow C$ such that $\psi \circ k = f \circ i \circ h$.But now, how can I find the unique $R-$module homomorphism $g:A \rightarrow C$ ? – TrItOs Dec 21 '23 at 19:43

1 Answers1

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Solution(1): Since $B$ is projective $R-$module, then there is a free $R-$module $F$ such that $F = B \oplus M$ where $M$ is $R-$module.

Since $B/A$ is projective $R-$module, then every short exact sequence $0 \rightarrow \cdot \rightarrow \cdot \rightarrow B/A \rightarrow 0$ splits.

Consider the short exact sequence $0 \rightarrow A \rightarrow B \rightarrow B/A \rightarrow 0$,

because $i : A \rightarrow B$ is injective $R-$module, $i(a)=a, \forall a \in A$ and $\pi : B \rightarrow B/A$ is surjective $R-$module, $\pi(b)=b+A, \forall b \in B$,

then this short exact sequence splits, i.e. $B = i(A) \oplus N/A = A \oplus N/A$ where $A$ is $R-$submodule of $N$ and $N$ is $R-$submodule of $B$.

Therefore $F = B \oplus M = A \oplus N/A \oplus M$ where $F$ is free $R-$module and $N/A \oplus M$ is $R-$module, then $A$ is projective $R-$module.

Solution(2): Since $B$ is projective $R-$module, then there is a free $R-$module $F$ such that $F = B \oplus M$ where $M$ is $R-$module.

Since $B/A$ is projective $R-$module, then there is a free $R-$module $L$ such that $L = B/A \oplus N$ where $N$ is $R-$module.

Therefore $\displaystyle{F/L = \frac{B \oplus M}{B/A \oplus N} \cong \frac{B}{B/A} \oplus \frac{M}{N} \cong A \oplus M/N}$ where $F/L$ is free $R-$module and $M/N$ is $R-$module, then $A$ is projective $R-$module.

Am I correct?

Which of the above solutions are correct?

TrItOs
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