# Advanced Modern Algebra, Ex 6.25

Been trying to figure this out for a while now (at least a week). The Problem Set was already handed in but I’m still trying to figure this out since I wasn’t able to answer this prior to handing it in. I really want to understand this better. Here’s the problem.

Let $0\to A\to B\to C\to 0$ be a short exact sequence of Modules. If $M$ is any module, prove that there are exact sequences.

$0\to A\oplus M\to B\oplus M\to C\to 0$

and

$0\to A\to B\oplus M\to C\oplus M\to 0$.

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"Adjoining a direct sum to a short exact sequence" or something similar would be a better title. –  AD. Oct 29 '12 at 16:45

Let $f:A\to B$ and $g:B\to C$ be the maps in your exact sequence above. Then let $\hat{f}:A\to B\oplus M$ be given by sending $a$ to $(f(a),0)$ and $\hat{g}:B\oplus M \to C\oplus M$ send $(b,m)$ to $(g(b),m)$. I will leave it to you now to show that $0\to A\to B\oplus M\to C\oplus M\to 0$ is exact.

The other problem is similar, so I will also leave it to you, but please comment if you get stuck.

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Suppose

$$0\rightarrow A\stackrel{f}\rightarrow B\stackrel{g}\rightarrow C\rightarrow 0$$

is exact, and let

$$0\rightarrow A\oplus M\stackrel{f\oplus i}\rightarrow B\oplus M\stackrel{\overline g}\rightarrow C\rightarrow 0$$

with

$$f\oplus i(a,m):=(f(a),m)\;\;,\;\;\overline g(b,m):=g(b)$$

then

$$\overline g\circ f(a,m)=\overline g(f(a),m):=gf(a)=0\Longrightarrow \operatorname {Im}f\subset \ker g$$

$$\overline g(b,m)=g(b)=0\Longrightarrow b\in\ker g=\operatorname {im}f\Longrightarrow \text{there exists}\;\; a\in A\,\,\text{s.t.}\,\,f(a)=b$$

$$\Longrightarrow (b,m)=(f(a),m)=f\oplus i(a,m)\Longrightarrow \ker g\subset\operatorname{Im}f$$

Since $\,f\oplus i\,$ is clearly $\,1-1\,$ and $\,\overline g\,$ is clearly onto, we're done.

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@ Brett Frankel You wanted to write $\hat g: B\oplus M \to C$ defined as $(b,m)\mapsto g(b)$ and not $(b,m)\mapsto (g(b), m)$. I wanted to write this as a comment. Sorry I am posting it as an answer. Can somebody please tell me how can I write a comment?
Actually, I was constructing the second exact sequence in your question. So the map is written correctly, except that I specified the wrong codomain. I should have written $\hat{g}:B\oplus M\to C\oplus M$. I'll make the correction to my solution. –  Brett Frankel Oct 29 '12 at 17:20