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:

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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?

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    $\begingroup$ You can take the normalisation of $\mathbf P^2$ in the larger field. $\endgroup$ Aug 12, 2020 at 17:29
  • $\begingroup$ Can you explain me please? $\endgroup$ Aug 12, 2020 at 18:03
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    $\begingroup$ 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. $\endgroup$ Aug 12, 2020 at 18:04
  • $\begingroup$ I know the concept of normalization but I don't understand the phrase 'in the larger field' $\endgroup$ Aug 12, 2020 at 18:06
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    $\begingroup$ 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 $\mathcal O_L$.) $\endgroup$ Aug 12, 2020 at 18:31

2 Answers 2


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 $\ell_1,\ldots,\ell_k$.

  • $\begingroup$ The problem is that the group $(Z/nZ)^k$ acts on $X$ and not $(Z/nZ)^{k-1}$. What is the mistake? $\endgroup$ Sep 5, 2020 at 15:03
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    $\begingroup$ 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. $\endgroup$ Sep 5, 2020 at 19:03
  • $\begingroup$ 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 $(a_1,..,a_k)\to (a_2-a_1, \cdots , a_k-a_1)$ ? $\endgroup$ Sep 6, 2020 at 10:07
  • $\begingroup$ Can you help me also for this question, please? math.stackexchange.com/q/3815141 $\endgroup$ 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 $(\mathbb{Z}/2\mathbb{Z})^2$.

  • $\begingroup$ But in this case $\pi$ would be a $(Z/2Z)^3$ covering and not a $(Z/2Z)^2$-covering, right? $\endgroup$ Sep 5, 2020 at 15:06
  • $\begingroup$ 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 $[z_0 \, : \,z_1] \mapsto[z_0^2 \, : \,z_1^2]$. $\endgroup$ Sep 5, 2020 at 18:50
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    $\begingroup$ This is double and not bi-double, for instance because $[z_0:z_1]=[-z_0: -z_1]$ and $[-z_0:z_1]=[z_0:-z_1]$. $\endgroup$ Sep 5, 2020 at 18:57

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