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This might be a stupid question... I‘m not good at Algebraic Geometry...

I'm reading a paper on Algebraic geometry code. There is a family of curve named as 'Garcia-Stichtenoth Curves'.

The function field for $X_1$ is $k(x_1)$, and for $X_2$ is $k(x_1,z_2)$.

[By the way, is it true that: for a smooth projective irreducible curve defined over $k$, expressed by a single equation $C(x,y)$ in $x,y$, the function field is $k(x,y)$, where $x,y$ are related by the equation $C$?]

The author said the function field for $X_3$ is $k(X_2)(z_3)$. So it equals $k(x_1,z_2,z_3)$. How to see that?

[It seems $k(X_3) = k(z_3,x_2)$, but $x_1$ is missing, this is where I feel confused].

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ZWJ
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1 Answers1

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It is true that the function field of a smooth projective irreducible curve $X$ defined over $k$ given by an equation $C(x,y)$. Recall that the function field of a curve $X$ is given by $\text{Frac}\left( k[x,y]/\langle C(x,y) \rangle \right)=k(x,y)$ with $C(x,y)=0$.

The idea in the Garcia-Stichtenoth curves and its function fields is that each function field is defined as an extension of the previous the same equation, that is, there exists a polynomial $f(Y,Z) \in k[Y,Z]$ such that $F_{i+1}=F_i(z_{i+1})$ with $f(z_i, z_{i+1})=0$. So $F_{i+1}=k(z_1, z_2, \dots, z_{i+1})$ and $f(z_j, z_{j+1})=0$ for all $1 \le j \le i$ (in your case, $z_1=x_1$).

In Garcia-Stichtenoth tower of function fields, we start with the rational function field $F_1:=k(x_1)$ over $k$ and we define $F_2=(x_1, z_2)$ by adjuncting a variable $z_2$ satisfying $z_2^{q_0}+z_2=x_1^{q_0+1}$. Thus, we obtain a finite extension $F_2/F_1$. Now, we repeat this to obtain $F_3$ from $F_2$, that is, $F_3:=k(x_1, z_2, z_3)$ with $z_3^{q_0} + z_3 = x_2^{q_0+1}=\left( \frac{z_2}{x_1} \right)^{q_0+1}$. And so on.

In fact, this is equivalent to constructing a sequence of curves $$ \cdots \to X_4 \to X_3 \to X_2 \to X_1 $$ where $X_i=Z(x_1, z_2^q+z_2-x_2^{q_0+1}, \dots, z_i^q+z_i-x_{i-1}^{q_0+1})$ and each morphism $X_{i+1} \to X_i$ is surjective.

Adler Marques
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