Let $R$ be a commutative ring with unity. Let $0\to M\xrightarrow{f} N\xrightarrow{g} P\to 0$ be a short exact sequence of $R$-modules.
I am looking at a proof of the fact that $N$ is Noetherian $\iff$ $M$ and $P$ are Noetherian.
For the forward direction, the proof states:
Since $f$ is an injection, every $R$-submodule of $M$ can be seen as an $R$-submodule of $N$.
Why is this true?
I have tried to think of a counterexample (and failed, probably because the statement is true), but this result does not seem obvious to me at all.
What I've tried:
I want to show that if $I$ is an $R$-submodule of $M$, then $f(I)$ is an $R$-submodule of $N$. That is, I want to show that $f(I)$ is an abelian group under $+_N$ (which I will just call $+$) and there is a well-defined ring action $\cdot : R\times f(I) \to f(I)$ satisfying the module axioms.
i) Since $R$-module homomorphisms are abelian group homomorphisms, and those preserve abelian groups, $f(I)$ is an abelian group.
ii) Since $f$ is an $R$-module homomorphism, we can define $r\cdot f(i)=f(r\cdot i)\in f(I)$. Then $$r\cdot (f(i)+f(j)) = r\cdot f(i+j) = f((r\cdot i)+(r\cdot j))= f(r\cdot i)+f(r\cdot j)=r\cdot f(i)+r\cdot f(j),$$ $$(r+s)\cdot f(i) = f((r+s)\cdot i) = f(r\cdot i)+(s\cdot i)) = r\cdot f(i)+s\cdot f(i),$$ and I can show other module axioms similarly.
My Questions:
a) Is this approach correct? Are my proofs correct?
b) Is there a good intuition behind this? For example, there is clearly an injection from $\mathbb{Z}$ to $\mathbb{Q}$, but no such injection preserves the ideals of $\mathbb{Z}$. Why should I take for granted that an injection will preserve submodules?