Suppose I am given a morphism of schemes $f: X \longrightarrow Y$ and suppose that for any point $y \in Y$, $\kappa(y)$ denotes the residue field. I would like to show that the fibered product $$X \times_{Y} \text{Spec}(\kappa(y))$$ is homeomorphic to the preimage $f^{-1}(y)$ in the induced topology.
The following is my attempt so far:
The question is obviously affine on the target, so without loss of generality we reduce to the case of a morphism $f: X \longrightarrow \text{Spec }A$. Then we have
$$
f^{-1}(\text{Spec }A) = \bigcup_{i \in I} \text{Spec }B_{i}
$$
via morphisms
$$
\phi_{i}: A \longrightarrow B_{i}.
$$
We can then write,
\begin{align*}
f^{-1}(\text{Spec } A) \times_{\text{Spec } A}\text{Spec } \kappa(y) &= \bigcup_{i \in I} \left( \text{Spec } B_{i} \times_{\text{Spec }A} \text{Spec }\kappa(y) \right) \\
&= \bigcup_{i \in I} \text{Spec } \left( B_{i} \otimes_{A} \kappa(y) \right).
\end{align*}
Now let $\mathfrak{p}$ be the prime ideal of $A$ corresponding to the point $y \in \text{Spec } A$. Then $\kappa(y) = (A/ \mathfrak{p})_{\mathfrak{p}}$. Then for any of the $B$ (dropping the index for brevity), we have
\begin{align*}
B\otimes_{A} (A/ \mathfrak{p})_{\mathfrak{p}} &\simeq B \otimes_{A} A / \mathfrak{p} \otimes_{A} A_{\mathfrak{p}} \\
&\simeq B_{\mathfrak{p}} \otimes_{A} A / \mathfrak{p} \\
& \simeq B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}}
\end{align*}
Now the prime ideals in $B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$. Further, the prime ideals of $B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B$ contained in $\phi(\mathfrak{p})B$. That is, we have
\begin{align}
\text{Spec }B_{\mathfrak{p}} &= \{ \mathfrak{q} \subseteq B : \phi^{-1}(\mathfrak{q}) \subseteq \mathfrak{p} \} \\
&= \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p} \}.
\end{align}
Note that in the very last line we have abused notation slightly: There is not really any notion of being "inside" $\mathfrak{p}$ when it is being treated as a point in the scheme. So really we mean that $f(\mathfrak{q})$ is a point in the scheme whose corresponding prime ideal is a subset of $\mathfrak{p}$. Now as we said above, the prime ideals of $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ are the following
$$
\text{Spec }(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}) = \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p} \text{ and } ??? \}
$$
This is the point at which I get stuck. Ideally I would like to make some argument from the statement "Now the prime ideals in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$" to conclude that the question marks are the condition $\mathfrak{p} \subseteq f(\mathfrak{q})$. Could someone point me in the right direction as to how to get this condition?
Is the proof up until this point correct? Different sources online seem to do this many different ways, so it's difficult for me to check.
Best Answer
The primes in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ are the primes in $B_{\mathfrak{p}}$ containing $\phi(\mathfrak{p})B_{\mathfrak{p}}$. And the primes in $B_{\mathfrak{p}}$ are the primes in $B$ not meeting $S:=\phi(A\setminus \mathfrak{p})$. Thus, a prime in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ is of the form $\mathfrak{q}B_{\mathfrak{p}}$ with $\mathfrak{q}\subset B$ not meeting $S$ such that $\phi(\mathfrak{p})B_{\mathfrak{p}}\subset \mathfrak{q}B_{\mathfrak{p}}$. Contracting yields $\phi(\mathfrak{p})\subset \mathfrak{q}$. On the other hand, $$ \mathfrak{q}\subset B\setminus \phi(A\setminus \mathfrak{p}). $$ Applying $\phi^{-1}$ and using $\phi(\mathfrak{p})\subset \mathfrak{q}$ yields $$ \mathfrak{p}\subset \phi^{-1}(\mathfrak{q}) \subset \phi^{-1}(B\setminus \phi(A\setminus \mathfrak{p})) \subset \mathfrak{p} $$ and consequently $\phi^{-1}(\mathfrak{q})=\mathfrak{p}$.