Fiber over a prime ideal Let $\phi: R \to S$ a ring morphism, $f: X = \mathrm{Spec}(S) \to \mathrm{Spec}(R) = Y$ the induced morphism, and $y := y_p \in Y$ the corresponding point to a prime ideal $p$ in $R$. 
How can I see that the topological fiber $f^{-1}(y)$ is bijective to $\mathrm{Spec}(S \otimes_R k(p))$ where $k(p) = \mathrm{Frac}(R/(p))$?
 A: Hint:
Consider the commutative diagram:
$$\begin{matrix}
R/\mathfrak p&\hspace-1em\longrightarrow\hspace-1em&\hspace-1emS/\mathfrak p S\\
\downarrow&&\hspace-1em\downarrow \\
\kappa(\mathfrak p)&\hspace-1em\longrightarrow\hspace-1em& S\otimes_R\kappa(\mathfrak p)
\end{matrix}$$
and the corresponding fiber product diagram for the spectra.
A: First of all you note that if $\mathfrak q$ contracts to $\mathfrak p$, we clearly have $\mathfrak q \supset \mathfrak pS$, i.e. the fibre $f^{-1}(x)$ is contained in $\operatorname{Spec}(S/\mathfrak pS)$. Thus you can pass to the map $$\operatorname{Spec}(S/\mathfrak pS) \to \operatorname{Spec}(R/\mathfrak p)$$
induced by the ring map $$R/\mathfrak p \to S/\mathfrak pS.$$
This shows that we only have to deal with the generic fibre, i.e. you have to show that for a ring map $R \to S$ from an integral domain $R$, the set of primes in $S$ that contract to zero can be identified with the primes of $S \otimes_R \operatorname{Frac}(R)$.
That is really automatic: By the well known behavior of prime ideals under localization, we know that prime ideals of said tensor product are just prime ideals of $S$, that do not meet $R \setminus \{0\}$, i.e. prime ideals that are contracted to zero under the given ring map.
A: A new attempt using hints (a bit formally):
According to the category equivalence of rings and affine schemes I can translate Bernard's diagram to:
$$\begin{matrix}
\mathrm{Spec}(S\otimes_R\kappa(\mathfrak p))&\hspace-1em\longrightarrow\hspace-1em&\mathrm{Spec}(\kappa(\mathfrak p))\\
\downarrow&&\hspace-1em\downarrow \\
\mathrm{Spec}(S/\mathfrak p S)&\hspace-1em\longrightarrow\hspace-1em& \mathrm{Spec}(R/\mathfrak p)
\end{matrix}$$
And because $S\otimes_R\kappa(\mathfrak p)$ is a push out the category equivalence again provides $\mathrm{Spec}(S\otimes_R\kappa(\mathfrak p)) \cong \mathrm{Spec}(S/\mathfrak p S) \times_{\mathrm{Spec}(R/\mathfrak p)} \mathrm{Spec}(\kappa(\mathfrak p))$ is a pull back. 
So according to the definition of a pull back we have $\mathrm{Spec}(S\otimes_R\kappa(\mathfrak p)) =\{p_S, p_{\kappa(\mathfrak p)}\mid p_S $ and  $p_{\kappa(\mathfrak p)}$ have the same image in the commut. quadrat $\} \cong \{(p_S, 0_{\kappa(\mathfrak p)} | \bar{f}(p_S) = \bar{0} \} \cong \bar{f}^{-1}(\bar{0})$ where $\bar{f}: \mathrm{Spec}(S/\mathfrak p S) \to \mathrm{Spec}(R/\mathfrak p)$ induced by $f$ and obviously according Moos' explanation we can identify $f^{-1}(p)$ with  $\bar{f}^{-1}(\bar{0})$. Is it ok?
