Prove that $\psi:\text{spec}(R_p)\to\text{spec} R$ is injective and $\text{Im}\psi=\{q\in\text{spec} R:q\subseteq p\}$

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Let $R$ be a ring with a unit and let $p$ be a prime ideal in $R$. Let $\phi:R\to R_p$ be the natural mapping $r\to {r\over 1}"="r$. Prove that the natural mapping $\psi:\text{spec}(R_p)\to\text{spec}(R),u\to\phi^{-1}[u]$ is injective, and that $\text{Im}\psi=\{q\in\text{spec}(R):q\subseteq p\}=:A$.

I showed that $\psi (u)$ is an ideal in $S$ but I didn't success to show that it is prime;

Let us define $S:=R_p$. We know that $S/u$ is an integral domain and we would like to show that $R/\psi(u)$ is an integral domain. Let $r+\psi(u)\ne\psi(u)$. Then $r\notin \psi (u)\Rightarrow r\notin u$. $$ \Rightarrow \exists a,b\in R,{a\over b}\cdot r+u=1+u \\\Rightarrow \exists 0\ne t\in R,0=t(ar-b) $$ I'm not sure how to proceed.