Baer sum of two extensions

Question

As we all know, all extensions of $Z/p$ by $Z/p$ are split extensions and following extensions:$$\epsilon_{i}:0\rightarrow Z/p\rightarrow Z/p^{2} \xrightarrow{i} Z/p \rightarrow0$$. Now the question is that how can we find the Baer sum of two extensions of $\epsilon_{i}$ and $\epsilon_{j}$ without using the fact that $Ext^{1}_{Z}(Z/p,Z/p)$ is isomorphic to the extensions with Baer sum.

The natural guess is that $\epsilon_{i}+\epsilon_{j}=\epsilon_{i+j}$ when $i+j\neq 0$. In order to prove this, we have to show there is a proper map from $Z/p^{2}\rightarrow X$ where $X$ is the extension group of the baer sum of $\epsilon_{i}$ and $\epsilon_{j}$ such that these two extensions are equivalent . Since from the definition of Baer sum, the structure of this $X$ is not so clear. How can we find such kind map?

Answer 1

Unfortunately your guess is the wrong one. Note that there is another natural option : $\epsilon_i+\epsilon_j=\epsilon_{(i^{-1}+j^{-1})^{-1}}$, or, using $\eta_i=\epsilon_{i^{-1}}$, then $\eta_i+\eta_j=\eta_{i+j}$.

(One way to see that there is probably inverse popping up is because the formula should not work if $i,j$ or $i+j$ are not invertible).

So the Baer sum $\eta_i+\eta_j$ is $X=U/V$ where : $$ U=\{ (a,b)\in \mathbb{Z}/p^2\mathbb{Z}\times\mathbb{Z}/p^2\mathbb{Z}~|~ i^{-1}a=j^{-1}b~\operatorname{mod}p\}$$

$$V=\{ (pa,-pa)\in \mathbb{Z}/p^2\mathbb{Z}\times\mathbb{Z}/p^2\mathbb{Z}\} $$

Let $u=i(i+j)^{-1}$ and $v=j(i+j)^{-1}$ where the inverse is taken in $\mathbb{Z}/p^2\mathbb{Z}$. Note that $u+v=1$ and $i^{-1}u=j^{-1}v$. Define a map $\mathbb{Z}/p^2\mathbb{Z}\to X$ by the formula : $a\mapsto (ua,va)$. This is well defined and we have a commutative diagram : $$\require{AMScd} \begin{CD} 0@>>>\mathbb{Z}/p\mathbb{Z}@>p>>\mathbb{Z}/p^2\mathbb{Z}@>(i+j) ^{-1}>>\mathbb{Z}/p\mathbb{Z}@>>>0\\ @.@|@VVV@|@.\\ 0@>>>\mathbb{Z}/p\mathbb{Z}@>>a\mapsto (pa,0)>X@>>(a,b)\mapsto i^{-1}a>\mathbb{Z}/p\mathbb{Z}@>>>0\\ \end{CD} $$ The above row is $\eta_{i+j}$. The diagram is commutative, indeed, this is obvious for the right square. For the left :

• the map $\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p ^2\mathbb{Z}\to X$ is $1\mapsto p\mapsto (up,vp)$
• the map $\mathbb{Z}/p\mathbb{Z}=\mathbb{Z}/p\mathbb{Z}\to X$ is $1\mapsto (p,0)$

But in $X$, $(up,vp)=(up+vp,0)=(p,0)$. This proves the claim.