@BalarkaSen So, in order to prove that $\epsilon_p$ is a ring homomorphismcould we do it like that? Or have I done something wrong?
Let $a,b \in \mathbb{Z}$. Then:
$\epsilon_p(a+b)=((a+b) \mod {p},(a+b) \mod {p^2}, (a+b) \mod {p^3}, \dots )=(a \mod{p}+b \mod{p}, a \mod{p^2}+b \mod{p^2}, a \mod{p^3}+b \mod{p^3}, \dots)=\epsilon_p(a)+ \epsilon_p(b)$
$\epsilon_p(ab)=(ab \mod {p},ab \mod {p^2}, ab \mod {p^3}, \dots )=(a \mod{p} \cdot b \mod{p}, a \mod{p^2} \cdot b \mod{p^2}, a \mod{p^3} \cdot b \mod{p^3}, \dots)=\epsilon_p(a) \cdot \epsilon_p(b)$