If every homomorphic image of an injective module is also injective, then every submodule of a projective module is projective

Question

Let $R$ be a ring with $1$. All modules considered in this problem are unitary right $R$-module.

Assume that every homomorphic image of an injective module is also injective. I need to prove that every submodule of a projective module is also projective.

How to prove this statement?! I need any hint. I tried to prove it by using the definition of the projectivity: Let $M$ be projective and $N\leq M$ any submodule. We need to prove that $N$ is projective. Let $\phi:B\to C$ be any epimorphism of modules and $\theta:N\to C$. We need to find $\psi:N\to B$ such that $\phi \psi= \theta$.

Answer 1

You can easily conclude that from the following characterization of projectivity.

A module $P$ is projective if (and only if) every diagram like the one below, with $E$ injective, can be completed. $$ \require{AMScd} \begin{CD} @. P \\ @. @VVV \\ E @>>> Q @>>> 0 \end{CD} $$

To prove this, I’ll use the well-known fact that every module embeds in an injective module. So, suppose that the above holds, and start with the usual diagram: $$ \require{AMScd} \begin{CD} @. P \\ @. @VV{\theta}V \\ B @>>{\phi}> C @>>> 0. \end{CD} $$ We want to find $\psi \colon P \to B$ such that $\phi\psi = \theta$.

• First, $B$ embeds in an injective module: There is a monomorphism $j \colon B \to E$ with $E$ injective.
• Next, if $i \colon \ker \phi \to B$ is the inclusion, and $\pi \colon E \to E/\operatorname{im}(ji)$ the canonical projection, consider the map $k \colon C \to E/\operatorname{im}(ji)$ defined as follows: Given $c \in C$, take $b \in B$ with $\phi(b)=c$ and define $k(c)$ as $\pi(j(b))$. Then the square below commutes. $$\require{AMScd} \begin{CD} @. P \\ @. @VV{\theta}V \\ B @>{\phi}>> C @>>> 0 \\ @V {j} VV @VV{k}V \\ E @>>{\pi}> {E/\operatorname{im}(ji)} @>>> 0 \end{CD} $$
• By assumption, there is some $h \colon P \to E$ such that $\pi h = k \theta$. Notice that for each $p \in P$ we have that $\theta(p) = \phi(b)$ for some $b \in B$; then applying $k$ we get $$\pi(h(p)) = \pi(j(b))$$ which means that $h(p)-j(b) \in \operatorname{im}(ji)$, and so $h(p) \in \operatorname{im} j$.
• Hence $\operatorname{im} h \subseteq \operatorname{im} j$, and then we can define $\psi \colon P \to B$ as follows: Given $p \in P$, choose $b \in B$ with $h(p)=j(b)$, and define $\psi(p)$ to be this $b$.