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I have confused with the definition of "Birational Equivalence" in the Algebraic Geometry.

In My Text book , ($V$ and $W$ are irreducible quasi-projective varieties)

A rational map $f : V \to W$ is called birational if there is a rational map $g : W \to V$ with

$f\circ g = \text{id}_W$ and $g\circ f=\text{id}_V$.


My question is the domain of $\text{id}_W$ and $\text{id}_V$.

How do we think the domain of these rational map?

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  • $\begingroup$ What are your own thoughts on what it might mean? $\endgroup$ – Joonas Ilmavirta Nov 4 '14 at 8:55
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A rational map does not have a domain. A rational map is an equivalence class of regular maps: Let $V$ be a variety. Then a rational map is an equivalence class of pairs $(\phi_U,U)$ where $\phi:U \to W$ is a regular map and $U subset V$ is an open subset. We say that two such pairs are equivalent if $\phi_U = \phi_{U'}$ on $U \cap U'$.

With this definition in mind, we say that two rational maps are equal if they agree on a dense open set. Thus the domain of $id_W, id_V$ are the whole of $W,V$, but the equality sign means "equal as rational maps".

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  • $\begingroup$ sorry for the late reply. thank you! $\endgroup$ – 8Frog Oct 29 '15 at 5:39

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