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Categorical constructions in the category of schemes usually do not preserve the underlying sets. For example, the underlying topological space of the product of schemes is not the topological product of the underlying spaces. It is true that if $V,W$ are varieties then the closed points of the product is the same as the product of the closed points, but even then the topology is finer than the product topology.

However, the scheme theoretic fiber over a point has the same underlying set and the same topology as the fiber in the category of toppological spaces. I know how to prove this result - but why should we expect this to be true?

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up vote 6 down vote accepted

Instead of giving a vague a-priori reason or the usual proof, let me give a quite unknown direct proof, which even works in the larger category of locally ringed spaces:

If $X \to S$ and $Y \to S$ are morphisms of locally ringed spaces, then you can construct their fiber product $X \times_S Y$ explicitly, see here for a summary. The set consists of all $(x,y,s,\mathfrak{p})$, where $x \in X, y \in Y$ lie over $s \in S$ and $\mathfrak{p}$ is a prime ideal of $\kappa(x) \otimes_{\kappa(s)} \kappa(y)$. The topology is generated by sets of the form $\{(x,y,s,\mathfrak{p}) : x \in U, y \in V, s \in W, \alpha(x,y) \notin \mathfrak{p}\}$ for $U \subseteq X, V \subseteq Y$ open which map into $W \subseteq S$ open, and $\alpha \in \mathcal{O}_X(U) \otimes_{\mathcal{O}_S(W)} \mathcal{O}_Y(V)$. The structure sheaf is defined in such a way that the stalk at $(x,y,s,\mathfrak{p})$ is the localization of $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$ at the preimage of $\mathfrak{p}$.

Low let $s \in S$ and consider $Y=\mathrm{Spec}(k(s))$ with the canonical morphism $Y \to S$. Let $f : X \to S$ be arbitrary. As a set, we have $Y=\{s\}$, with structure sheaf $\mathcal{O}_{Y,s}=\kappa(s)$. Given a point $(x,y,s,\mathfrak{p})$ in $X \times_S Y$, we then have $y=s$ and $\mathfrak{p}$ is a prime ideal in $\kappa(x) \otimes_{\kappa(s)} \kappa(s)=\kappa(x)$, i.e. $\mathfrak{p}=0$. Thus, $X \times_S Y \cong f^{-1}(s)$ as sets. The topology is generated by sets of the form $\{x \in f^{-1}(s) : x \in U, r(x,s) \neq 0\}$ for $U \subseteq X$, $s \in W \subseteq S$ open and $r \in \mathcal{O}_X(U) \otimes_{\mathcal{O}_S(W)} \kappa(s)$. But these sets are open in the usual topology of $f^{-1}(s)$ (and the converse is clear from $r=1$): If $r$ as above satisfies $r(x,s) \neq 0$ in $\kappa(x)$, then $r_x \in \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \kappa(s)$ is invertible, but the inverse extends to an open neighborhood of $x$ and we see that there is an open subset $x \in U' \subseteq U$ such that $r(x',s) \neq 0$ for all $x' \in U'$.

Edit. The underlying set $|X|$ of a locally ringed space $X$ (or just scheme) can be recovered from its $k$-valued points where $k$ runs through all fields as follows: $$|X| \cong \mathrm{colim}_k ~X(k)$$ From this and $(X \times_S Y)(k) = X(k) \times_{S(k)} Y(k)$ (which is the universal property of the fiber product) one easily deduces the description of $|X \times_S Y|$ given above.

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Dear Martin, This is a nice answer. I think your edit hits the key point. Indeed, when I saw the question I was going to write an answer explaining exactly this, and then saw that you had already done so in your answer! Cheers, – Matt E Dec 15 '13 at 21:47
Yes this explains easily how the underlying set of the LRS-fiber looks like. But I don't know how to obtain the topology from the universal property without an explicit construction. – Martin Brandenburg Dec 15 '13 at 21:50

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