Question about fibered products of schemes (Liu exercise 3.1.7) I'm having a bit of difficulty writing out the details for exercise 3.1.7 in Liu's 'Algebraic Geometry and Arithmetic Curves'. The question is as follows:
Let $X,Y$ be $S$-schemes, $p$ and $q$ be the projection morphisms from $X\times_{S}Y$ to $X$ and $Y$ respectively, and fix $s \in S$. Show that for any $x \in X_{s} := X \times_{S} Spec(k(s))$ and $y \in Y_{s}$ there exists a natural homeomorphism
$$Spec(k(x) \otimes_{k(s)} k(y)) \to \{z \in X \times_{S} Y| p(z) = x, q(z) =y\}$$
Some thoughts: I know we can rewrite the left hand side as
$$Spec(k(x) \otimes_{k(s)} k(y)) \simeq Spec(k(x)) \times_{Spec(k(s))} Spec(k(y)) \simeq Spec(k(x)) \times_{S} Spec(k(y))$$
from which we obtain a map into $\{z \in X\times_{S}Y| p(z) = x, q(z) = y\}$ via the universal property of fiber products. Also, we may rewrite the set as
$$\{z \in X\times_{S}Y| p(z) = x, q(z) = y\} = p^{-1}(x) \cap q^{-1}(y) = (Y \times_{S} Spec(k(x))) \cap (X \times_{S} Spec(k(y))).$$
Finally, I know that the fiber over a point is homeomorphic to the preimage. At this point I'm not sure what direction to go, and any hints are greatly appreciated. 
 A: Let $x,y,s$ are chosen and fixed. Note that the problem only involves neighbourhoods of the points $x,y,s$ and is independent of the whole scheme $X,Y,S$. So one first chooses an affine open neighbourhood of $s$ say $spec(C)$, then choose open affine neighbourhoods of $x,y$ say $spec(A), spec(B)$ respectively, mapping into $spec(C)$ via $\pi_X, \pi_Y$ in that order. Let $x = \mathfrak{p}_x, y = \mathfrak{p}_y, s = \mathfrak{p}_s$ be the corresponding prime ideals in the rings $A,B,C$ respectively. Note that by defintion these prime ideals satisfy the relation $\pi_X^{-1}(\mathfrak{p}_x) = \pi_Y^{-1}(\mathfrak{p}_y) = \mathfrak{p}_s$. Let $pr_X : spec(A\otimes_CB) \rightarrow spec(A), pr_Y : spec(A\otimes_CB) \rightarrow spec(B)$. Then
$Z = \lbrace \mathfrak{p} \in spec(A\otimes_CB)| pr_X^{-1}(\mathfrak{p}) = \mathfrak{p}_x, pr_Y^{-1}(\mathfrak{p}) = \mathfrak{p}_y \rbrace$ = 
$\lbrace \mathfrak{p} \in spec(A/\mathfrak{p}_sA)\otimes_{(C/ \mathfrak{p}_s)}(B/\mathfrak{p}_sB) | \overline{pr_X}^{-1}(\mathfrak{p}) = \overline{\mathfrak{p}}_x, \overline{pr_Y}^{-1}(\mathfrak{p}) = \overline{\mathfrak{p}}_y\rbrace$
since prime ideals in $X,Y$ restrict to same prime ideals in $S$. This is a bijection seems okay. Note that $(A/\mathfrak{p}_sA)\otimes_{(C/ \mathfrak{p}_s)}(B/\mathfrak{p}_sB) \cong (A\otimes_C B)\otimes_C (C/\mathfrak{p}_s)$. On the level of spectra this is the same as considering inverse image of the point $\mathfrak{p}_s$ under the map $spec(A\otimes_CB) \rightarrow C$, and hence thee sets are homeomorphism as the set in the first line is the inverse image as topological spaces and the space in the second line is a scheme theoretic inverse image.
$ = \lbrace \mathfrak{p} \in spec(A/\mathfrak{p}_sA)\otimes_{k(s)}(B/\mathfrak{p}_sB) | \overline{pr_X}^{-1}(\mathfrak{p}) = \overline{\mathfrak{p}}_x, \overline{pr_Y}^{-1}(\mathfrak{p}) = \overline{\mathfrak{p}}_y \rbrace = \lbrace \mathfrak{p} \in spec(A/\mathfrak{p}_sA)\otimes_{k(s)}(B/\mathfrak{p}_sB) | \mathfrak{p} \supset \overline{\mathfrak{p}}_x, \mathfrak{p} \supset \overline{\mathfrak{p}}_y \rbrace = \lbrace \mathfrak{p} \in spec(A/\mathfrak{p}_sA)\otimes_{k(s)}(B/\mathfrak{p}_sB)/ (\mathfrak{p}_x\otimes B + A \otimes \mathfrak{p}_y) \rbrace =_{by ~  definition} \lbrace \mathfrak{p} \in spec((A/\mathfrak{p}_x) \otimes_{k(s)} (B/\mathfrak{p}_y)) | \mathfrak{p} \cap A/\mathfrak{p}_x = 0 = \mathfrak{p} \cap B/ \mathfrak{p}_y \rbrace$
here taking inverse image is written as intersection in the ring $((A/\mathfrak{p}_x) \otimes_{k(s)} (B/\mathfrak{p}_y))$ with the image of $A/\mathfrak{p}_x$ under $\overline{pr_X}$ which is just the map $a \rightarrow a\otimes 1$, similarily for the $Y$ part of the statement. Thus we can invert the elements of the type $A/\mathfrak{p}_x \otimes \overline{1}$ and $\overline{1} \otimes B/\mathfrak{p}_y$. We get that the above set equals
$ = \lbrace spec(k(x) \otimes_{k(s)} k(y)) \rbrace$.
which was what we wanted to show.
