# Short exact sequences and different extension

Let $A = \mathbb Z$ and $B = \mathbb Q.$ Then Ext$(A, B)$ gives the set of all equivalent extensions of $A$ by $B.$ I have few questions.

1. Is this sequence $0\rightarrow \mathbb Z \stackrel{\alpha}{\longrightarrow} \mathbb Z \oplus \mathbb Q \stackrel{\beta} {\longrightarrow} \mathbb Q \rightarrow 0$ split?

My answer is yes if $\beta$ is the canonical projection on second components. Since we can get splitting map $\delta: \mathbb Q \rightarrow \mathbb Z \oplus \mathbb Q$ defined by $\delta(a) = (0,a).$ This satisfies $\beta\circ\delta$ identity on $\mathbb Q.$

1. Suppose $\beta$ is not a projection. Is the sequence in (1) split or not? I do not know exactly if there exists $\beta$ other than projection so that the sequence (1) is short exact.

2. Suppose the above sequence in (1) is split and suppose we have another short exact sequence $0\rightarrow \mathbb Z \stackrel{\alpha}{\longrightarrow} G \stackrel{\beta} {\longrightarrow} \mathbb Q \rightarrow 0,$ which is equivalent with the sequence in $(1).$ Can we say $G\cong \mathbb Z\oplus\mathbb Q$? Thanks

• The sequence $(1)$ must be split since $\delta$ is the spliting map.
– CAA
Jul 12, 2015 at 22:15

Concerning the third question the answer is yes. Suppose that two short exact sequences are equivalent. Then the middle groups are isomorphic by using the $5$-lemma (the morphism between the middle groups is injective and surjective). The converse need not be true. An example here are the $p$ inequivalent extensions of $C_p$ by $C_p$, while there are only two non-isomorphic groups: $C_{p^2}$ and $C_p\times C_p$. So equivalence is finer than isomorphism in general. A detailed discussion can be found at this MSE question here.