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an homework question :

Let $X$ and $Y$ be two integral schemes of finite type over a field $k$. There is a rational morphism $X\longrightarrow Y$ if and only if the scheme $Y\times_k k(X)$ has a rational point ($k(X)$ denotes the function field of $X$).

Any idea on how to prove that?

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1 Answer 1

To go from the rational map to the rational point, try answering these two questions: what (regular) maps are there from $K(X)$ to anything obvious; and if you have a rational map from $X$ to $Y$, what regular maps to $Y$ is it related to?

To go from the rational point to the rational map, first try doing it when $X$ and $Y$ are both affine, which will show you why finite type is necessary. Then answer these questions: can $X$ be assumed to be affine; and can $Y$ be assumed to be affine?

Some further intuition might be helpful, though I don't know if it will actually aid in solving the problem. A regular map $X \to Y$ is determined by its graph, an $X$-valued point of $X \times_k Y$. This problem is asking you to show that a rational map is also determined by its "graph" in $K(X) \times_k Y$. Figuring out what that means is, in some sense, the proof.

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I'm sorry but i don't manage to get any further. Could you give me more details? –  Dude May 7 '11 at 10:14

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