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!
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?