Understaing the proof of genus of smooth projective curve.

Question

I'am reading Miranda's book that proves that the genus $g(X)$ of a smooth projective complex plane curve $X=\left\{[x: y: z] \in \mathbb{P}^2 \mid F(x, y, z)=0\right\}$ of degree $d$ is equal to $$ g(X)=\frac{(d-1)(d-2)}{2} $$ He takes the standard projection $$ \begin{gathered} \pi: \mathbb{P}^2 \rightarrow \mathbb{P}^1 \\ {[x: y: z] \rightarrow[x: z]} \end{gathered} $$ in his prove.

My quesion is that this projection is not well-defined usually, does he mean that under some linear trasformation we can make this projection well-defined ?

Let us consider a elliptic curve $Y^{2}Z=X(X-Z)(X-2Z)$, then [0:1:0] is on the curve hence $\pi$ is not well-defined.

If we replace x by x+cy and z by z+dy then we can avoid that!

Answer 1

You don't need to perform a linear transformation to dodge the point $[0 : 1 : 0]$.

It's true that the projection $\pi : \mathbb P^2 \to \mathbb P^1$ is only a rational map. However, if $C$ is a smooth curve in $\mathbb P^2$, then the restriction $\pi|_{C} : C \to \mathbb P^1$ is guaranteed to be a morphism.

I'm appealing to a general result, which says that every rational map from a smooth curve to a $\mathbb P^n$ is a morphism.

To illustrate this, let $C$ be the elliptic curve that you mentioned in your question. Let's focus on the affine patch $$ \mathbb A^2 \cong \{ [x : 1 : z] : (x, z) \in \mathbb A^2 \} \subset \mathbb P^2, $$ which includes the potentially problematic point $[0 : 1 : 0]$. In this affine patch, the curve $C$ is the vanishing locus of the polynomial $$ f(x,z) = z - x(x-z)(x - 2z).$$

The rational map $\pi|_C : C \to \mathbb P^1$ is the map $$ (x, z) \mapsto [x : z],$$ and naively, it looks like $\pi|_C$ is not regular at the point $p = (0, 0)$.

However, this rational map can be rewritten as $$ (x, z) \mapsto [1 : z/x]. $$ The gradient of $f$ at $(0, 0)$ is $$ \nabla f(p) = \left( \frac{\partial f}{\partial x}(p), \frac{\partial f}{\partial z}(p)\right) = ( 0, 1).$$ This gradient is non-vanishing, so $C$ is smooth at $p$. Furthermore, the vector $(1, 0)$ complements the vector $\nabla f(p) = (0, 1)$ to form a basis for the two-dimensional vector space $k^2$. So $x$ is a local parameter at $p$, i.e. $x$ generates the unique maximal ideal in the local ring at $p$.

The local ring at $p$ is a discrete valuation ring. Let $v_p$ be the valuation.

• $z$ vanishes at $p$, so $v_p(z) \geq 1$.
• $x$ is a local parameter at $p$, so $v_p(x) = 1$.

Hence $v_p(z / x) = v_p(z) - v_p(x) \geq 0$. This means that $z / x$ is regular at $p$.

So the map $\pi|_C$, which sends $ (x, z) \mapsto [1 : z/x]$, is regular at $p$.

[Alternatively, you can use the equation $z = x(x-z)(x-2z)$ to write $$ z/x = (x-z)(x-2z).$$ So $\pi|_C : C \to \mathbb P^2$ can be written as $$ (x, z) \mapsto [1 : z/x] = [1 : (x - z)(x-2z)],$$ which is manifestly regular at $p$.]

Finally, I'll explain how the technique generalises. Suppose $C$ is any smooth curve in $\mathbb A^2$ containing the point $p = (0, 0)$.

Let $t$ a local parameter at $p$. Let $a := \text{min}(v_p(x), v_p(z))$. Then $\pi|_C$ can be represented as

$$ (x, z) \mapsto [t^{-a} x : t^{-a} z].$$

Notice that:

• $v_p(t^{-a}x) \geq 0$ and $v_p(t^{-a}z) \geq 0$, so both $t^{-a}x$ and $t^{-a}z$ are regular at $p$.
• Either $v_p(t^{-a}x) = 0$ or $v_p(t^{-a}z) = 0$, so at least one of $t^{-a}x$ or $t^{-a}z$ is non-vanishing at $p$.

Therefore, the map $ (x, z) \mapsto [t^{-a} x : t^{-a} z]$ is manifestly regular at $p$, which is to say that $\pi|_C$ is regular at $p$.