Let $0 \to A \stackrel{i}{\to} B \stackrel{p}{\to} C \to 0$ be a short exact sequence of $R$-modules.
Suppose that $A = \langle X \rangle$ and $C = \langle Y \rangle$
For each $y \in C$, choose y' \in B such that p(y')=y. Prove that B = \langle i(x) \cup \{ y':y \in Y \} \rangle
I just can't seem to figure out how to get this to work - I don't think it should be hard!
Let $a \in A$, then $a = \sum r_i x_i$. Similarly c = \sum r'_i y_i = \sum r'_i p(y'_i).
Obivously I should use exactness: $\operatorname{img} i = \operatorname{ker} p$
$b \in \operatorname{img} i \implies b = \sum r_i i(x_i)$
The $p(b) = 0$ gives that $pi(x_i)=0$ - but that is just clear from the definitions!
I am thinking that to be in the kernel of $p$, we must have p(y'_i)=0
Any hints to point me in the right direction?