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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?

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Let $b\in B$, write $p(b) = \sum_k r_k y_k$ (a minor remark: it is a bad idea to use the same letter for indices in a sum and for a homomorphism, like you did in your post). Now, wouldn't you just love to be able to claim that b = \sum_k r_k y'_k. Of course that's not true, but what do you know about the difference b - \sum_k r_k y'_k? You should be able to take it from here.

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    thanks for that. (I hadn't even really noticed I was using $i$ for the homomorphism and the indicies!$)2011-06-07
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You don't want to start with $a\in A$; rather, you should start with $b\in B$ (and try to show that $b$ can be expressed using the $\iota(x)$ and the $y'$).

To that end, note that you can certainly use the y' to construct an element b' of $B$ that has the same image under $p$ as $b$; this because the images of the y' generate $C$.

But if $b$ and b' have the same image under $p$, then b-b'\in \mathrm{ker}(p) = \mathrm{Im}(\iota); so you can express b-b' in terms of $\iota(X)$.

So: b-b' can be expressed in terms of $\iota(X)$, and b' can be expressed in terms of the y'. That means that $b$ can be expressed in terms of $\iota(X)$ and the y', which proves that $B$ is contained in the submodule generated by the given set.