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Let $V \subset \mathbb{A}^n$ be an affine variety, and $\varphi: V \to V$ a birational map. Does there exist a unique birational map $\varphi' : V' \to V'$ of the projective closure $V' \subset \mathbb {P}^n$ of $V$ that agrees with $\varphi$ on $V \subset V'$?

A rational map is an equivalence class of morphisms $U \to V$ with $U \subset V$ nonempty open, where two are equivalent if they agree on the intersection of their domains.

I'm thinking the answer is affirmative because $V \subset V'$ is a nonempty open subset under the usual embedding, and a rational map is uniquely determined by any single representative.

If $\varphi$ is locally given in coordinates by rational functions, how does one compute $\varphi'$? I'm thinking one can clear the denominators and homogenize to arrive at a map of the projective closure locally given by homogeneous polynomials of the same degree (which locally gives a morphism).

If necessary: I'm most interested in the case $n=2$ where $V$ is a curve $Z(f)$ for an irreducible $f$, so $V' = Z(f^*)$ where $f^*$ is the homogenization of $f$.

Community
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Ricardo Buring
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1 Answers1

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Of course, a birational map from V to V is the same as an isomorphism from a dense open subset U of V to another dense open subset f(U). And now if V' is the closure of V, then both U and f(U) are dense in V', so f is a birational map from V' to V'.

lee
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