Algebra extension of modules question

Question

Let $M$ be an extension of modules of $A$ by $B$ (modules), i.e. $A \leqslant M$ and $M/A \cong B$. Show that if $M$ is finitely generated then so is $B$.

In the solution it just says: If $M = XR$ then $B = π(X)R$ where $π : M → B$ is the quotient map - is this intuitive? I can't see it. Any help greatly appreciated!

Answer 1

Let $M=XR$ which means that every elements can be written by $R$ combination of $X$.

Now, Let $b\in B$ then there exist an $m\in M$ such that $\pi(m)=b$. We know that $$m=r_1x_1+...+r_nx_n$$ $$b=\pi(m)=r_1\pi(x_1)+...r_n\pi(x_n)$$

So we see that any element of $b$ can be written as $R$ combiantion of $\pi(X)$.

Thus, $B=\pi(X)R$ and by finiteness of $X$, we are done.