The variety induced by an extension of a field

Question

If $K$ is a finitely generated field extension of $k$, then there exists an irreducible affine $k$-variety with function field $K$. The idea is that if $x_1, \dots, x_n$ are generators of $K$ under $k$, i.e each elements of $K$ is a rational function in $x_1, \dots , x_n$, then the kernel of the map $k[t_1,\dots, t_n]\to K$ is a prime ideal and the induced map between their field fractions is an isomorphism:

$(k[t_1,\dots, t_n]/I)_0\cong K$

This means $Z(I)\subseteq k^n$ is the affine irreducible variety which field fraction corresponds to $K$.

Now I have the following problem:

In this case I have $k$ equal to the function field of $\mathbb{P}^2$, and $K$ equal to the finite extension $k((\frac{l_2}{l_1})^{\frac{1}{n}},\dots, , (\frac{l_k}{l_1})^{\frac{1}{n}})$. In the paper the author tells us $K$ determine an algebraic (affine?) surface $X$ with normal singularities and a natural map $\pi: X\to \mathbb{P}^2$.

I don't understand how to define this natural map $\pi$ and what is exactly this surface $X$. I think that $K$ determine an affine variety up to birational morphisms and so I don't understand how to define exactly $X$.

Can you give me an example for $n=2$ and $k=3$, please?

Answer 1

I decided to turn my comment into an answer not because it is complete but because I think it can be of use.

Let $z=(z_0:z_1:z_2)$ and $u=(u_1:\ldots:u_k)$ be homogeneous coordinates of $\mathbb{P}^2$ and $\mathbb{P}^{k-1}$. First notice that the surface $X_1\subset \mathbb{P}^2\times \mathbb{P}^{k-1}$ defined by the vanishing of the $2\times 2$-minors of the matrix \begin{equation*} \begin{pmatrix} u_{1} & u_{2} & \cdots & u_{k} \\ \ell_{1} & \ell_{2} & \cdots & \ell_{k} \\ \end{pmatrix} \end{equation*} is the closure of the graph of the rational map $z\mapsto (\ell_1:\ldots:\ell_k)$. Restricting the projection you get a well defined map $X_1\rightarrow \mathbb{P}^2$.

On the other hand you also have a $n$-to-$1$ map $\mathbb{P}^{k-1}\rightarrow \mathbb{P}^{k-1}$ given by $\phi_n:(t_1:\ldots:t_k)\mapsto (t^n_1:\ldots:t^n_k)$. This induces $id\times\phi_n:\mathbb{P}^2\times\mathbb{P}^{k-1}\rightarrow \mathbb{P}^2\times\mathbb{P}^{k-1}$. Now you can take $X$ to be the preimage of $X_1$ by $id\times\phi_n$.

In this way you can "see" $X\subset \mathbb{P}^2\times\mathbb{P}^{k-1}$ with coordinates $(z,t)$ as the vanishing set of minors of the matrix

\begin{equation*} \begin{pmatrix} t^n_{1} & t^n_{2} & \cdots & t^n_{k} \\ \ell_{1} & \ell_{2} & \cdots & \ell_{k} \\ \end{pmatrix}. \end{equation*}

Also the map $\pi:X\rightarrow \mathbb{P}^2$ is clear. The ramification locus is induced by the ramification locus of $\phi_n$.

I'm not sure about the singularities of $X$ but I think they will depend on the relative position of lines $\ell_1,\ldots,\ell_k$.

Answer 2

Let $n=2$ and $k=3$, and suppose by the sake of simplicity that the three lines are in general position. Then, up to projective transformations, we can assume that they are the three coordinate lines $\ell_1$, $\ell_2$, $\ell_3$ given by $z_0=0$, $z_1=0$, $z_2=0$, respectively.

Then your function field is simply $\mathbb{C}(x, \, y)(\sqrt{x}, \, \sqrt{y})$, where $x=z_1/z_0$, $y=z_2/z_0$, and the affine equation of your $(\mathbb{Z}/2\mathbb{Z})^2$-cover $X \to \mathbb{P}^2$ on the chart $z_0 \neq 0$ is $$(x, \, y) \mapsto (x^2, \, y^2).$$

Note that $X$ is projective, since it is a finite covering of a projective variety; in fact, $X$ is the union of three of these affine charts, corresponding to the three standard charts for $\mathbb{P}^2$.

A moment of thought shows that $X = \mathbb{P}^2$, and that the global equation of your bi-double cover is $$\pi \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad [z_0: \, z_1: \, z_2] \mapsto [z_0^2: \, z_1^2: \, z_2^2].$$

It is an instructive exercise to factor $\pi$ through the three singular double covers $$X_i \to \mathbb{P}^2, \quad i=1,\, 2, \, 3$$ corresponding to the three non-trivial involutions in the Klein group $(\mathbb{Z}/2\mathbb{Z})^2$.