# Short exact sequence - Why doesn't this argument work?

What is wrong with this "proof"?

If the sequence of $\Bbb{Z}$-modules $$0\to M \to N \to \Bbb{Z}/2 \to 0$$ is exact, then $N\cong M \oplus \Bbb{Z}/2$.

Call the first map $f$, the second $g$. By exactness, we have \begin{aligned} \Bbb{Z}/2 &= \operatorname{Im} g\\ \ker g &= \operatorname{Im} f \\ \ker f &= 0. \end{aligned}

By the first isomorphism theorem, we have $$\Bbb{Z}/2 \cong N / \ker g = N/\operatorname{Im} f\\ \operatorname{Im} f \cong M/ \ker f = M.$$

So $\Bbb{Z}/2 \cong N/M$, i.e. $N \cong M \oplus \Bbb{Z}/2$.

• Your last step is incorrect. – user99914 May 19 '15 at 9:25
• $N = \mathbb Z/4\mathbb Z$, $M=\mathbb Z/2\mathbb Z$ also satisfies the last line. – Mathmo123 May 19 '15 at 9:29

The exact sequence may not be split. $N/M\cong\mathbb{Z}_2$ does not imply that $N\cong M\oplus\mathbb{Z}_2$. A counterexample is $N\cong M\cong\mathbb{Z}$ and $M\to N$ is the multiplication by 2 map.

• If we consider this exact sequence instead $$0\to M \to N \to \Bbb{Z} \to 0$$ then the $N/M \cong \Bbb{Z}$ does imply that $N\cong M\oplus \Bbb{Z}$, right? How can we justify that? – iwriteonbananas May 19 '15 at 10:10
• Yes. We can always lift the isomorphism $\mathbb{Z}\stackrel{\cong}{\to}N/M$ to an injective homomorphism $j: \mathbb{Z}\hookrightarrow M$. The map $(i, j): M\oplus\mathbb{Z}\to N$ is the desired isomorphism. – Alex Fok May 19 '15 at 11:23

The condition

for every short exact sequence of $R$-modules $0\to M\to N\to P\to 0$, we have $N\cong M\oplus P$

is equivalent to saying that $P$ is a projective $R$-module.

Of course $\mathbb{Z}/2\mathbb{Z}$ is not projective, because it can't be embedded in a free module to begin with.

Take $0 \rightarrow \mathbb{Z}_p \rightarrow \mathbb{Z}/p^2\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$ where the first map is $\overline{a} \rightarrow pa \,( \mbox{ mod } p^2)$ and the second map is $x + py \in \mathbb{Z}/p^2\mathbb{Z} \rightarrow x$. Now this is an exact sequence. But $\mathbb{Z}/p^2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}_p \oplus\mathbb{Z}_p$. This is because $M/N \equiv Q$ does not necessarily mean $M \equiv N \oplus Q$.