# Finding the regular points of a rational map

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. – Qiaochu Yuan Feb 27 '11 at 17:28
The restricted morphism is from $L$ to $\phi(L)$ and not, as you wrote , the other way round. – Georges Elencwajg Jul 15 '13 at 11:46

a) The restriction of $\phi$ to any curve $C$ disjoint from $P$ is a regular morphism $C\to \mathbb P^2$.
b) The restriction of $\phi$ to any smooth curve $S$ is a regular morphism $S\to \mathbb P^2$, even if $S$ contains a point of $P$.
This is a very important property of smooth curves : rational maps from them into projective varieties are actually regular, i.e. are morphisms.
Note carefully that this is completely false for singular curves or smooth varieties of dimension $\geq 2$.

For your example of the line $L$, both a) and b) apply and you needn't "fiddle about with polynomials and equations" !

References
Hartshorne, Chapter I, Proposition 6.8, page 43 .
or
Shafarevich, Volume 1, Chapter II, §3, Corollary 1 to Theorem 3, page 110 .

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Pretty late ,I know, but as for your question about the domain of the "standard quadratic transformation", you might be interested in the following question and its answer. Basically, you consider the induced morphism \begin{split} \phi_{cone}: \Bbb{A}^3 &\rightarrow \Bbb{A}^3\\ (x,y,z) & \mapsto (yz,zx,xy) \end{split} and deduce that every extension of $\phi$ to $\Bbb{P}^2$ would have to map $(1:0:0),(0:1:0),(0:0:1)$ to the "point" $(0:0:0)$, which is of course impossible.

Also, there is a general criterion for deciding whether or not a point is in the domain of some rational map, based on the relation of "dominance" between local rings (cf. Fulton, Proposition 6.6.11).

Proposition: Let $F: X \dashrightarrow Y$ be a dominating rational map between varieties. Then a point $P \in X$ belongs to the domain of $F$ and $F(P)=Q$, if and only if the local ring $\mathcal{O}_P(X)$ dominates the local ring $\tilde{F}(\mathcal{O}_Q(Y))$, where $\tilde{F}: k(Y) \rightarrow k(X)$ denotes the induced embedding of function fields.

However, I am not sure if this result is too helpful in practice.

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