# Short exact sequence of modules generated by a set

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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## 2 Answers

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!$) – Juan S Jun 7 '11 at 3:07 You don't want to start with$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.

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