How can one show that the a rational map $f:V\rightarrow W$ is Zariski-continuous? ($V$ and $W$ are affine varieties, i.e. irreducible closed algebraic sets.)
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Assaf, i see you're new here, so: did you know you can "accept" an answer by clicking the check mark next to it? :) This shows that you are happy with the answer.. – Joachim Apr 08 '13 at 23:16
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Perhaps you wanted to use the term "regular function" or "morphism of varieties" instead of "rational map". (To the future readers of this question.) – Patrick Da Silva May 23 '17 at 23:45
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Since a rational map $V\dashrightarrow W$ is technically not a map $V\to W$, I hope it suffices to show that for some open $U\subseteq V$ where $f$ is defined, $f_U:U\to W$ is continuous. For that, it will suffice to show that morphisms of algebraic varieties are continuous. Well then, let $Z\subseteq W$ be closed. Let $f:U\to W$ be a morphism of algebraic varieties and let $\phi:k[W]\to k[U]$ be the corresponding morphism of coordinate rings, i.e. $\phi(a)= a\circ f$. Note that, if $p\in U$ corresponds to the maximal ideal $I(p)$, then $f(p)$ corresponds to the maximal ideal $\phi^{-1}(I(p))$. \begin{align*} f^{-1}(Z) &= \{ p\in U\mid f(p)\in Z \} \\ &= \{p\in U\mid I(Z) \subseteq I(f(p)) = \phi^{-1}(I(p)) \} \\ &= \{p\in U\mid \phi(I(Z)) \subseteq I(p) \} \\ &= Z(\phi(I(Z))). \end{align*} Hence, the preimage of a closed set is closed.
Jesko Hüttenhain
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