The variety induced by an extension of a field

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?

• You can take the normalisation of $\mathbf P^2$ in the larger field. Aug 12, 2020 at 17:29
• Can you explain me please? Aug 12, 2020 at 18:03
• One place you can read about relative normalisation is Tag 035H. I don't know what your background is, so it's hard to say more. Aug 12, 2020 at 18:04
• I know the concept of normalization but I don't understand the phrase 'in the larger field' Aug 12, 2020 at 18:06
• You have a finite field extension $K \to L$ and a normal (even smooth) variety $X$ with fraction field $K$. Then you can take the normalisation of $X$ in $L$, meaning on each affine open $U = \operatorname{Spec} A$ (so $\operatorname{Frac} A = K$) you take the integral closure of $A$ in $L$, and glue these together for the various $U \subseteq X$. (The cited tag is merely a coordinate-free way to phrase this: the pushforward of $\mathcal O_L$ is a quasi-coherent $\mathcal O_X$-module, and you take the relative Spec of the integral closure of $\mathcal O_X$ in $\mathcal O_L$.) Aug 12, 2020 at 18:31

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 $$\ell_1,\ldots,\ell_k$$.

• The problem is that the group $(Z/nZ)^k$ acts on $X$ and not $(Z/nZ)^{k-1}$. What is the mistake? Sep 5, 2020 at 15:03
• The group acting effectively is $(Z/nZ)^{k-1}$, because the element $(-1,\ldots,-1) \in (Z/nZ)^k$ acts trivially on the homogeneous coordinates. Sep 5, 2020 at 19:03
• Yes, ok, but this would mean simply that the elements of the type $(a,...,a)$ belongs to the stabilizer of each point, right? We are interested only of that actions that are faithful, so this mean we must consider the new faithful action $((Z/nZ)^k)/(\cap_x G_x)$, that is isomorphic exactly to (Z/nZ)^{k-1} via the isomorphism $(a_1,..,a_k)\to (a_2-a_1, \cdots , a_k-a_1)$ ? Sep 6, 2020 at 10:07
• Can you help me also for this question, please? math.stackexchange.com/q/3815141 Sep 6, 2020 at 10:10

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 $$(\mathbb{Z}/2\mathbb{Z})^2$$.

• But in this case $\pi$ would be a $(Z/2Z)^3$ covering and not a $(Z/2Z)^2$-covering, right? Sep 5, 2020 at 15:06
• If $k=3$ then $k-1=2$. The covering in the affine chart is clearly a bi-double cover, non tri-double. So the extension to the projective plane must be bi-double, too. Recall that $z_0$, $z_1$, $z_2$ are homogeneous coordinates. Think of the double cover given in affine coordinates by $x \mapsto x^2$: in homogeneous coordinates it becomes $[z_0 \, : \,z_1] \mapsto[z_0^2 \, : \,z_1^2]$. Sep 5, 2020 at 18:50
• This is double and not bi-double, for instance because $[z_0:z_1]=[-z_0: -z_1]$ and $[-z_0:z_1]=[z_0:-z_1]$. Sep 5, 2020 at 18:57