Extensions of $\mathbb{Z}_p$ by $\mathbb{Z}$ (Hilton & Stammbach III.1.2)

Question

Question is to compute $E(\mathbb{Z}_p,\mathbb{Z})$ i.e., equivalence classes of extensions of $\mathbb{Z}_p$ by $\mathbb{Z}$

By an extension of $A$ by $B$ i mean an $R$ module $E$ such that $B\rightarrow E\rightarrow A$ is an exact sequence.

I say $B\rightarrow E_1\rightarrow A$ and $B\rightarrow E_2\rightarrow A$ are equivalent if there exists a homomorphism $\zeta : E_1\rightarrow E_2$ such that the following diagram commutes

\begin{matrix} B&\to&E_1&\to&A\\ \|&&\;\;\downarrow f&&\|\\ B&\to&E_2&\to&A \end{matrix}

Now First question is to show that :

$$\mathbb{Z}\xrightarrow {\mu} \mathbb{Z}\xrightarrow{\epsilon} \mathbb{Z}_3 ~\text{and }~ \mathbb{Z}\xrightarrow {\mu'} \mathbb{Z}\xrightarrow{\epsilon'} \mathbb{Z}_3 $$ where $\mu=\mu'$ is multiplication by $3$, $\epsilon(1)=1 (\mod 3)$ and $\epsilon'(1)=2 (\mod 3)$ are not equivalent.

Second Question is (use above idea) to compute $E(\mathbb{Z}_p,\mathbb{Z})$...

I have solved first exercise and with that i realized that there are atleast $p-1$ non equivalent extensions with $\mu$ being multiplication by $p$ and $\epsilon_i=i\mod p$ for $1\leq i\leq p-1$.

And then I have split extension $$\mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/p\mathbb{Z}$$

So i have $p$ extensions now..

I am not able to prove there are no other extensions which are not equivalent to any of this..

Please help me to see this....

If any one can make that commutative diagram using latex i will be so thankful.. I tried to google for drawing commutative diagrams but i am not able to do as of now..

Thank you.

This is an exercise in Hilton & Stammbach's A Course in Homological algebra chapter $3$, Exercise $1.2$

Answer 1

Suppose $\def\ZZ{\mathbb Z}0\to\ZZ\xrightarrow{f}E\xrightarrow{g}\ZZ_p\to0$ is a non-split extension.

If $E$ has non-zero torsion, then that torsion clearly does not intersect the image of $f$, and therefore is mapped injectively into $\ZZ_p$. You can easily construct, then, a map $\ZZ_p\to E$ which is a section to $g$: this contradicts non-splitness.

It follows that $E$ is torsion free. As Jyrki observes, $E$ is either cyclic or a direct sum of two cyclic groups. It it were a direct sum of two cyclic groups, its quotient by the image of $f$ would be infinite. It follows that $E$ is isomorphic to $\ZZ$. Now there are exactly $p-1$ surjective homomorphisms from $\ZZ$ to $\ZZ_p$, etc.

Answer 2

Since this is supposed to be homological algebra, let's compute $\operatorname{Ext}^1(\mathbb{Z}/p,\mathbb{Z})$.

We have the exact sequence $0\to \mathbb{Z}\xrightarrow{\cdot p}\mathbb{Z}\to\mathbb{Z}/p\to 0$. Applying $\operatorname{Hom}(\mathbb{Z}/p,-)$, we see that $0 \to \mathbb{Z}/p\to \operatorname{Ext}^1(\mathbb{Z}/p,\mathbb{Z})\to \operatorname{Ext}^1(\mathbb{Z}/p,\mathbb{Z}/p)$ is exact. But $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/p,\mathbb{Z}/p)=\operatorname{Ext}^1_{\mathbb{Z}/p}(\mathbb{Z}/p,\mathbb{Z}/p) = 0$ ($\mathbb{Z}/p$ is a projective module over itself), so $\operatorname{Ext}^1(\mathbb{Z}/p,\mathbb{Z}) \cong \mathbb{Z}/p$, i.e. there are exactly $p$ inequivalent extensions.