Regularity of the rational map $[X:Y:Z]\mapsto[X^2:XY:Z^2]$?

Question

How to see that the mapping $[X:Y:Z]\mapsto[X^2:XY:Z^2]$ from the variety $V(Y^2Z-X^3-Z^3)\subset\mathbb{P}^2$ to $\mathbb{P}^2$ is regular at $[0:1:0]$?

Answer 1

Note that you can write $Z = \frac{X^3}{Y^2-Z^2}$ and by plugging this expression in you can factor out the $X$.

In particular by doing this, you will see that the rational map $[X:Y:Z] \mapsto [X(Y^2-Z^2)^2: Y(Y^2-Z^2)^2: X^5]$ defined on $Y\ne \pm Z$ agrees with the original rational map on the common locus on which they are defined, and this rational map is defined at $[0:1:0]$.

In general the locus of indeterminacy of a rational map from a non-singular variety $X$ to $\mathbb{P}^m$ has codimension $\ge 2$ (see for example Shafarevich Theorem 2.12), so you can always extend a rational map to projective space to a morphism in the case of a non-singular curve.

Answer 2

As @loch notes, a rational map from a non-singular curve to projective space will always be defined everywhere. It can often be helpful to work through examples by following the proof of the more general statement, so let's do that (in much detail):

The proof of the statement for curves involves taking a uniformizer $t$ for the point $P$ on the curve $C$, and then dividing out by $\text{ord}_P (f_i)$ where $f_i$ are the equations that define the morphism $\phi$.

So let's go through that with this curve in particular. The maximal ideal corresponding to the point $P=[0:1:0]$ is generated by $(X, Z) = \mathfrak{m}_P \subset k[X,Y,Z]_P$.

We should expect this ideal to be principal (because the curve is non-singular -- check!), and indeed since $Z = \frac{X^3}{Y^2 - Z^2} = X \cdot \frac{X^2}{Y^2-Z^2}$ we see that $Z$ is in the principal ideal generated by $(X)$, so $\mathfrak{m}_P = (X)$. From this, we also see that $Z$ has valuation $3$ (since it has a zero of order 3 from $X^3$, and $Y^2 - Z^2$ has a non-zero value at the point $[0:1:0]$).

So the trick is that $[X^2 : XY : Z^2]$ has terms of valuation $2, 1, $ and $6$ respectively. By re-writing $Z^2$, we can factor out one extra factor of $X$, and see that this map is indeed regular at $[0:1:0]$. Written out:

$[X^2 : XY : Z^2] = [X^2 : XY : \frac{X^6}{(Z^2-Y^2)^2}] = [X(Z^2-Y^2)^2 : Y(Z^2-Y^2)^2 : X^5]$