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Let $X$ and $Y$ be (irreducible, quasi-projective) varieties (over an algebraically closed field $k$), and $\phi: X \to Y$ be a rational map. I think I understand what it means for $\phi$ to be regular at a point $x \in X$, but I'm lost as to how to prove that $x$ is (or isn't) a regular point of $\phi$, when I am given a single representative of $\phi$.

For example, suppose $X = Y = \mathbb{P}^2$ is the projective plane, and $\phi$ is the rational map determined by $\phi([x : y: z]) = [yz : zx : xy]$. This uniquely defines a rational map since $\mathbb{P}^2$ is irreducible and the equation makes sense away from $P = \{ [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1] \}$, which is a closed set. I can see that $\phi$ is a birational equivalence (and even an involution!), and it seems intuitive that $\phi$ cannot be regular on $P$. But how do I prove this rigorously?

Then, for a twist, consider the line $L \subset \mathbb{P}^2$ which satisfies the equation $x + y + z = 0$. Clearly, $L \cap P = \emptyset$, and $\phi(L)$ is the conic $xy + yz + zx = 0$, and $P \subset \phi(L)$. It turns out $\phi$ restricted to $\phi(L)$ is a morphism of varieties $\phi(L) \to L$, because we've thrown away enough of the obstructions to regularisation. So regularity seems to be something a bit more subtle than staring at polynomials... but the way I showed that this restriction did turn $\phi$ into an honest morphism was precisely that: I fiddled about with polynomials and equations until I got a consistent set which covered the whole curve. Is there a better / more general way of finding the points at which a a rational map is regular?

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    Restrict to an open affine neighborhood and see if you can prove or disprove that the morphism restricts to a morphism of affines. At least, in principle.2011-02-27
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    The restricted morphism is from $L$ to $\phi(L)$ and not, as you wrote , the other way round.2013-07-15

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