# How to compute the topological space of fibered product of schemes?

I know that the topological space of fibered product of schemes is generally distinct to the usual Cartesian product of toplogical spaces of schemes. Then how can we compute the top. sp. of fibered product of sch. explicitly? Is there any systematic procedure that I can do?

Actually this question arises from when I read the Hartshorne's Algebraic Geometry. In section 4 on Chapter2, the example says that the affine line with doubled origin(I may denote it by X) is not separated over the field k. In the explanation of the book, I can see that the top. sp. of the fibered product of two X . However, I cannot understand how to compute it. (If you want the exact statement, see the p.96 of Hartshorne)

You can read the precise construction in Hartshorne's proof of the existence of the fiber product. Suppose we have $S$-schemes $X \xrightarrow{f} S$ and $Y \xrightarrow{g} S$. Let $S = \bigcup_i W_i$ be an affine open cover and also take affine open covers $f^{-1}(W_i) = \bigcup_{j} U_{ij}$ and $g^{-1}(W_i) = \bigcup_{j} V_{ij}$ for each $i$.  The fiber products $U_{ij} \times_{W_i} V_{ij}$ are just the spectrums of the tensor products of the coordinate rings. Now the fiber product $X \times_S Y$ is constructed by gluing these affine schemes, on certain open subsets, first to $f^{-1}(W_i) \times_{W_i} V_{ij}$, then to $f^{-1}(W_i) \times_{W_i} g^{-1}(W_i)$, and finally over the base to $X \times_S Y$. You can read about the gluing construction in (Marco Lo Guidice's notes, 2.3).

Basically, given schemes $X$ and $Y$ with open subsets $U \subset X$, $V \subset Y$ and an isomorphism $\phi : U \stackrel{\sim}{\to} V$, one constructs the underlying topological space of the gluing by taking the disjoint union of the spaces quotiented by the equivalence relation identifying $u \sim \phi(u)$ for each $u \in U$.

I happen to have a detailed construction of the fiber product of the affine line with double origin with itself that I wrote for homework at some point, here it is. I've just copied and pasted without proofreading again, so there may be typos, etc. $\newcommand{Spec}{\mathrm{Spec}\ }$

Consider the affine schemes $X_1 = \Spec k[x]$, $X_2 = \Spec k[y]$.  Between the open subsets $U_1 = \{ (x - a) : a \ne 0 \} \subset X_1$ and $U_2 = \{ (y - b) : b \ne 0 \} \subset X_2$ there is a natural isomorphism $\phi : U_1 \to U_2$.  Let $X$ be the scheme obtained by gluing $X_1$ and $X_2$ along $\phi$.  Topologically, its points are equivalence classes of the disjoint union of $X_1$ and $X_2$ by the relation $\sim$ under which points $(x - a)$ are identified with their images $(y - a)$, for $a \ne 0$.  Let $\bar{U_i} \subset X$ be the images of $U_i$ by the natural maps $X_i \to X$.  Note $\bar{U_1} \cap \bar{U_2} = \{ [(x - a)] : a \ne 0 \}$.

To compute the fiber product of $X$ with itself over $\Spec k$, we follow the construction of Hartshorne of the fiber product of general schemes: first we glue $X_1 \times_k X_1$ and $X_2 \times_k X_1$ by certain open sets to obtain $X \times_k X_1$, and we glue $X_1 \times_k X_2$ and $X_2 \times_k X_2$ to obtain $X \times_k X_2$, and then we glue these together to obtain $X \times_k X$.  We can compute $X_1 \times_k X_1 = \Spec k[x] \otimes k[x] \cong \Spec k[t_1, t_2]$ and $X_2 \times_k X_1 = \Spec k[y] \otimes k[x] \cong \Spec k[t_3, t_4]$.   Let $p_1$ be the first projection map associated with $X_1 \times_k X_1$.  Note $p_1$ maps the equivalence class $[(x-a)] = \{(x - a), (y - a)\} \in X$ to $(x - a) \in X_1$ and the class $[(0)] = \{(0)\}$ to $(0) \in X_1$.  Now consider the open set $U_1' = p_1^{-1} (\bar{U_1} \cap \bar{U_2}) = \{ (t_1 - a, t_2 - b) : a \ne 0, b \in k \}$.  Similarly let $U_2' = p_2^{-1} (\bar{U_1} \cap \bar{U_2}) = \{ (t_3 - a, t_4 - b) : a \ne 0, b \in k \}$, where $p_2$ is the first projection map associated with $X_2 \times_k X_1$.  Hartshorne shows that the result of gluing $X_1 \times_k X_1$ and $X_2 \times_k X_1$ via the natural isomorphism $\phi' : U_1' \to U_2'$ is the fiber product $X \times_k X_1$.  Topologically we find it is the disjoint union of $X_1 \times_k X_1$ and $X_2 \times_k X_1$ with $(t_1 - a, t_2 - b)$ identified with $(t_3 - a, t_4 - b)$ for all $a \ne 0, b \in k$.  Analogously we find $X \times_k X_2$ to be the disjoint union of $X_1 \times_k X_2$ and $X_2 \times_k X_2$ with $(t_1 - a, t_2 - b)$ identified with $(t_3 - a, t_4 - b)$ for all $a \in k, b \ne 0$.

Now we glue $X \times_k X_1$ and $X \times_k X_2$.  Let $q_1$ and $q_2$ be the second projection maps associated with $X \times_k X_1$ and $X \times_k X_2$, respectively.  Note that $q_1$ maps equivalence classes $[(t_1 - a, t_2 - b)]$ and $[(t_3 - a, t_4 - b)]$ to $(x - b) \in X_1$ and similarly $q_2$ maps them to $(y - b) \in X_2$.  Let $U_1'' = q_1^{-1} (\bar{U_1} \cap \bar{U_2})$.  This consists of equivalence classes $[(t_1 - a, t_2 - b)] = [(t_3 - a, t_4 - b)]$ with $a \ne 0$, $b \ne 0$, and classes $[(t_1, t_2 - b)]$, $[(t_3, t_4 - b)]$ with $b \ne 0$.  Similarly let $U_2'' = q_2^{-1} (\bar{U_1} \cap \bar{U_2})$ consists of the equivalence classes $[(t_1 - a, t_2 - b)] = [(t_3 - a, t_4 - b)]$ with $a \ne 0, b \ne 0$, along with $[(t_1 - a, t_2)]$ and $[(t_3 - a, t_4)]$ with $a \ne 0$.  We have an isomorphism $\phi'' : U_1'' \to U_2''$ which maps $[(t_1 - a, t_2 - b)] \mapsto [(t_1 - b, t_2 - a)]$ and $[(t_3 - a, t_4 - b)] \mapsto [(t_3 - b, t_4 - a)]$.  The result of gluing $X \times_k X_1$ and $X \times_k X_2$ via $\phi''$ is the fiber product $X \times_k X$.

• Many Thanks! It seems fairly complicate and annoying computation. Maybe, that is the reason why I hesitate to compute along with the proof of the existence of fiber product(as you know, there are many things such as finding top. sp. of subspace, gluing etc). – User0829 Mar 6 '13 at 7:05

If $X,Y$ are locally ringed spaces over a locally ringed space $S$, then one can write down their fiber product $X \times_S Y$ explicitly. If $X,Y,S$ are schemes, then this fiber product is a scheme, so that it is the fiber product in the category of schemes. This provides an alternative (and in my opinion far better) construction of the fiber product of schemes which is not well-known as it should be. You can find it here. Let me sketch it:

Let $f : X \to S$ and $g : Y \to S$. The underlying set of $X \times_S Y$ consists of those triples $(x,y,s,\mathfrak{p})$, where $(x,y,s)$ lies in the underlying topological fiber product, i.e. $x \in X, y \in Y, s \in S$ and $f(x)=s=g(y)$, and $\mathfrak{p}$ is a prime ideal in $\kappa(x) \otimes_{\kappa(s)} \kappa(y)$, the tensor product of the residue fields. In other words, the fiber of the map from $X \times_S Y$ to the topological fiber product is precisely $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(s)} \kappa(y))$ at the point $(x,y,s)$. Note that $\mathfrak{p}$ can also be equivalently described as a prime ideal $\mathfrak{q}$ in the tensor product of the local rings $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$, which restricts to the maximal ideals in $\mathcal{O}_{X,x}$ and $\mathcal{O}_{Y,y}$ (and therefore also in $\mathcal{O}_{S,s}$, since $f^\#_x$ and $g^\#_y$ are local).

The topology looks as follows: If $U \times_T V$ is a basic open subset of the topological fiber product (i.e. $U \subseteq X, V \subseteq Y, T \subseteq S$ open wih $f(U) \subseteq T \supseteq g(V)$) and $f \in \mathcal{O}_X(U) \otimes_{\mathcal{O}_S(T)} \mathcal{O}_Y(V)$, then $\Omega(U,V,T,f) \subseteq X \times_S Y$ consists of those $(x,y,s,\mathfrak{q})$ such that $x \in U, y \in V, s \in T$ and that the image of $f$ in $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$ is not contained in $\mathfrak{q}$. These sets form a basis for a topology on $X \times_S Y$. The structure sheaf is defined in such a way that on local rings $\mathcal{O}_{X \times_S Y,(x,y,s,\mathfrak{q})} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y})_{\mathfrak{q}}$. A general section in $\mathcal{O}_{X \times_S Y}$ is a function valued in these stalks, which is locally a fraction.

Of course, all this is not needed to see that the affine line with doubled origin is not separated. And once again I cannot recommend to read Hartshorne for the basics of algebraic geometry. If $X_1$ and $X_2$ are equal affine schemes and $U$ is an open subscheme, then $X=X_1 \cup_U X_2$ is separated iff $X_1 \times_X X_2 = X_1 \cap X_2 = U$ is affine, and $U \to X_1 \times X_2$ is a closed immersion. In particular, the affine line with a doubled origin is not separated.

• +1 for the Hartshorne comment. And this alternate construction of the fiber product looks very interesting, thanks for sharing. – user314 Mar 6 '13 at 10:05