Homomorphisms between $R/(p_1^{e_1})$ and $R/(p_2^{e_2})$ when $p_1$ and $p_2$ are associated.

Question

I have given that $R$ is a PID, $p_1,p_2 \in R$ are associate prime elements and $e_1,e_2 \in \mathbb{Z}_{>0}$. I now have to show that \begin{equation} \mathrm{Hom}_R\big(R/(p_1^{e_1}), R/(p_2^{e_2})\big) \cong R/(p_1^{\min(e_1,e_2)}). \end{equation}

Because $p_1$ and $p_2$ are associated, so $(p_2^{e_2}) = (p_1^{e_2})$. Moreover, we know that a morphism from \begin{equation} \varphi \colon R/(p_1^{e_1}) \to R/(p_1^{e_2}) \end{equation} is completely determined by $\varphi(\overline{1})$. However, I don't know how I can continue.

Answer 1

Hint:

Composing on the right by the canonical homomorphism $\pi:R\longrightarrow R/(p_1^{e_1})$, $\varphi$ corresponds to a homomorphism $R\longrightarrow R/(p_2^{e_2})$ which vanishes on the ideal $(p_1^{e_1})$ (universal property of kernels).