Fiber of a morphism of spectra Let $f:R\to S$ be a ring homomorphism, and for a fixed prime ideal  $\mathfrak p\subset R$, let $p:R\to k(\mathfrak p)$ be the canonical  homomorphism, $k(\mathfrak  p)$ being the fraction field  of $R/\mathfrak p$. Take the following (pushout) square: $\require{AMScd}$ $$\begin{CD}
R@>>p> k(\mathfrak p)\\
@VVfV @VjVV\\
S@>i>>S\otimes_R k(\mathfrak p)
\end{CD}$$ and consider its image (in $\rm Top$) under  the functor $\rm Spec$ (whose action I'll denote by $-^*$). So $p^*$ is injective, being its domain a singleton ($R\neq 0$), and also $i^*$ is injective: $i=l\circ q$, where $q:S\to S\otimes_R R/\mathfrak p$ is surjective, and $l: S\otimes _R R/\mathfrak p\to S\otimes_R  k(\mathfrak p)$ is a localization (respect to the subset of elements of the form $1\otimes \bar r$, $r\notin \mathfrak p$) hence $q^*,l^*$ are injective. Until now, we can say that $i^*$ injects its domain into the fiber of $\mathfrak p\in \operatorname{Spec} R$; but I don't understand how to deduce the most important part, i.e. that the image of $i^*$ is the entire fiber (and so the square in $\rm Top$ is a pullback). I apologize for the third question of the same kind in  two days, but this passage (that I found on Clark's Commutative Algebra, 4.3) is where my confusion actually started. Thanks for your patience
 A: Let's translate this back into commutative algebra. Explicitly you need to show that
$$\{\mathfrak{q}\in\operatorname{Spec}S\mid f^{-1}(\mathfrak{q}) = \mathfrak{p}\}\subseteq i^\ast(\operatorname{Spec}S\otimes_R k(\mathfrak{p}));$$
i.e., that for every prime ideal $\mathfrak{q}\subseteq S$ which pulls back to $\mathfrak{p}\subseteq R,$ there exists a prime ideal $\mathfrak{Q}\subseteq S\otimes_R k(\mathfrak{p})$ with $i^{-1}(\mathfrak{Q}) = \mathfrak{q}.$
Here are some hints to help you finish this argument; the complete details are below under the spoiler tags.
Hint 1: Break the problem down into showing that $\mathfrak{q}$ is the inverse image of some prime $\mathfrak{q}'\subseteq S\otimes_R R/\mathfrak{p}$, and that this $\mathfrak{q}'$ is the inverse image of a prime $\mathfrak{Q}\subseteq S\otimes_R k(\mathfrak{p}).$
Hint 2:

 Recall that if $A$ is a ring and if $I\subseteq A$ is an ideal of $A,$ then there is a bijection between prime ideals of $A/I$ and prime ideals of $A$ containing $I.$

Hint 3:

Recall also that if $A$ is a ring and $T\subseteq A$ is a multiplicative subset, that there is a bijection between prime ideals of $T^{-1}A$ and prime ideals of $A$ not meeting $T.$

Full solution:

Take $\mathfrak{q}\subseteq S$ such that $f^{-1}(\mathfrak{q})=\mathfrak{p}.$ We know that $f^{-1}(\mathfrak{q}) = \mathfrak{p},$ so that $\mathfrak{q}\supseteq\mathfrak{p}S.$ By hint 2, there exists a prime ideal $\mathfrak{q}'\subseteq S\otimes_R R/\mathfrak{p}$ with $q^{-1}(\mathfrak{q}') = \mathfrak{q}.$ The explicit form of the bijection mentioned tells us that $\mathfrak{q}' = q(\mathfrak{q}).$


Following hint 3 above, we must show that that $q(\mathfrak{q})\cap U = \emptyset.$ Observe that $$U = \{1\otimes\overline{r}\mid r\not\in\mathfrak{p}\} = \{f(r)\otimes 1\mid r\not\in\mathfrak{p}\} = \{q(f(r))\mid r\not\in\mathfrak{p}\}\subseteq (q\circ f)\left(R\setminus\mathfrak{p}\right).$$ Suppose for the sake of contradiction that there exists $r\in R$ such that $1\otimes \overline{r}\in q(\mathfrak{q})\cap U.$ Then there exists $s\in \mathfrak{q}$ such that $q(s) = q(f(r)).$ In other words, there would be an $s\in\mathfrak{q}\cap f(R\setminus\mathfrak{p}),$ which is impossible since $f^{-1}(\mathfrak{q}) = \mathfrak{p}.$ This completes the proof.

