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Let $ f : X \to S, g : Y \to S$ be the scheme morphisms, and $ X \times_S Y$ be the product of shemes. Let $ z \in X \times_S Y$ , and $ x=p(z) , y=q(z) ,$ where $ p: X \times_S Y \to X, q: X \times_S Y\to Y $ are projections. Suppose $ s=f(x)=g(y) \in S$ and $x\in U \subset X, y\in V \subset Y, s\in W \subset S, f(U) \subset W, g(V) \subset W$ where $U, V, W$ are arbitrary open sets. Is it ture $ z\in U \times_W V$ ?

(The question should be read as : Is it ture that $ z$ is in the image of open immersion $U \times_W V \to X \times_S Y$ )

Just a little comment: this can be used to prove monomorphism (particularly open and closed immersions) is seperated. The reason is, by the above result, one can pick up an open cover of $ X \times_Y X$ as $ U \times_V U$, because of the monomorphism property, two projections are the same, so one can choose the same open set $U$, and by the morphism of affines is seperated, we are done.

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up vote 2 down vote accepted

Let $Z=\{z\}$. Then $p$ and $q$ induce morphisms $Z\to U$ and $Z\to V$ compatible with the structure maps $U\to W$ and $V\to W$. Then by the definition of a product there exists a unique morphism $Z\to U\times_W V$ compatible with these morphisms. You can repeat the same argument for $X,Y,S$ and get a unique morphism compatible with the appropriate morphisms. This has to equal the one coming from the original embedding $Z\hookrightarrow X\times_SY$. Now clearly the first one combined with the natural morphism $U\times_W V\to X\times_S Y$ gives another such morphism so by uniqueness it has to be equal to the original embedding, so in other words the latter has to factor through $U\times_W V\to X\times_S Y$.

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Nice argument! It is essentially like the second approach in the second answer, but successfully avoid complicated argument. – Li Zhan Jun 2 '11 at 3:02

With the notations above, it can be easily checked (basically because open immersions are monomorphism; or more generally using that pullbacks commute with pullbcks), that $p^{-1}(U) \cap q^{-1}(V)$ satisfies the universal property of $U \times_W V$ (together with the obvious maps to $U$ and $V$) so that in particular the canonical morphism $U \times_W V \to X \times_S Y$ is isomorphic to the open immersion $p^{-1}(U) \cap q^{-1}(V) \to X \times_S Y$. Actually this basic argument is used in the gluing construction of the fiber product of schemes. I wonder how you can know that $U \times_W Y \to X \times_Y Y$ is an open immersion without this identification.

There is another, more concrete approach. Namely, the fiber product exists in the category of locally ringed spaces, can be written down explicitly, and turns out to be a scheme if we plug in schemes (see here). A point in $X \times_S Y$ is then a point $(x,y)$ over $s$ in the topological fiber product together with a prime ideal in $k(x) \otimes_{k(s)} k(y)$. Now you can also see directly that for $x \in U, y \in V$, this is just a point in $U \times_W V$.

Concerning your comment: In every category a monomorphism $X \to Y$ has the property that the diagonal $X \to X \times_Y X$ is an isomorphism. In particular, every monomorphism of schemes is separated. What you meant is probably that every injective morphism of schemes is separated.

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Thank you for your detailed explanation! I did not know that : In every category a monomorphism has the property that the diagonal is an isomorphism. Could you please say more about this? I know it boil down to prove the affine case (in the category of scheme), but I cannot prove this. As for the open immersion $U \times_W V$, it comes from the construction of product : the product is the gluing of shemes of the form $U \times_W V$, and by the gluing lemma, one knows each piece has an open immersion to the target scheme. – Li Zhan 0 secs ago – Li Zhan Jun 2 '11 at 3:24
If $X \to Y$ is a monomorphism, then two projections $X \times_Y X \to X$ coincide (since this is the case after applying $X \to Y$), and yield a morphism inverse to the diagonal. This argument works in every category. – Martin Brandenburg Jun 2 '11 at 8:47

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