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Let $S$ be a scheme. Let $Z = X\times_S Y$ be the product of $S$-schemes. Let $f \colon X \rightarrow S, g\colon Y \rightarrow S, h\colon Z \rightarrow S$ be the structure morphisms. Then $h(Z) = f(X) \cap g(Y)$?

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up vote 6 down vote accepted


Proof: Certainly, since the map $h$ factors through both $f$ and $g$, we have an inclusion $h(Z) \subset f(X) \cap g(Y)$. On the other hand, suppose that $s \in f(X) \cap g(Y).$ Then the fibre of $Z$ over $s$ is equal to the product of fibres $X_s$ and $Y_s$ over $\kappa(s)$. Each of these fibres is non-empty by assumption, and hence so is there fibre product.

Here is a more concrete argument (it is just a variation on the proof that a product of non-empty schemes over a field is non-empty): There exists $x \in X$ mapping to $s$, and also $y \in Y$ mapping to $s$, so we get maps of residue fields $\kappa(s) \hookrightarrow \kappa(x)$ and $\kappa(s) \hookrightarrow \kappa(y)$. Now choose any maximal ideal $\mathfrak m$ of $\kappa(x)\otimes_{\kappa(s)}\kappa(y)$, with residue field $\Omega,$ say. Then the maps $\kappa(x), \kappa(y) \to \Omega$ induce a map Spec $\Omega \to X \times_S Y$, and the image is a point $z \in Z$ such that $h(z) = s$. Thus we get the desired equality. QED

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