Weil’s theorem and its applications

An interesting problem is to determinate when a field of Moduli is a field of definition. The following result given by Weil [A. Weil, The field of definition of a variety. Amer. J. Math 78] will give us necessary and sufficient conditions for this.

Theorem: (Weil’s theorem)
Let $\small K$ < $\small F$ be a finite Galois extension,
$\small X\subset\mathbb{P}^n(F)$ a projective algebraic variety defined over $\small F$ and $\small K=M_{F/K}(F)$ . Suppose that for each $\small \sigma\in F_K(X)=Gal(F/K)$ exists a birational application $\small f_\sigma:X\to X^\sigma$ defined over $\small F$ such that, for each pair $\small \sigma,\tau\in Gal(F/K)$ , $\small f_{\sigma\tau}=f_\tau^\sigma\circ f_\sigma$ holds, that is the following diagram commutes
$\small \xymatrix{ X \ar[r]^{f_\sigma} \ar[d]_{f_{\sigma\tau}} & X^\sigma \ar[ld]^{f_\tau^\sigma}\\ X^{\sigma\tau} & }$
Then:
1.- There exists a projective algebraic variety $\small Y$ , defined over $\small K$ , and there exists a birational application $\small R:X\to Y$ defined over $\small F$ such that, for every $\small \sigma\in Gal(F/K)$ it holds the equality $\small R^\sigma\circ f_\sigma=R$ . Moreover, if every $\small f_\sigma$ is biregular, then we can assume $\small R$ to be a biregular isomorphism.
2.- If the pair $\small (\widehat{R},\widehat{Y})$ is another solution for the above problem, then exists a birational isomorphism $\small \varphi:Y\to\widehat{Y}$ defined over $\small F$ such that, $\small \widehat{R}=\varphi\circ R$

PROOF: of Weils’ theorem

Now, if $\small X$ is a non-singularprojective algebraic curve of genus $\small g\geq 2$ defined over a field $\small F$ , then its automorphism group $\small Aut (X)$ is finite. We can consider the Galois covering $\small P:X\to X/Aut(X))$ . And in this case,
$\small M_{F/K}(P:X\to X/Aut(X),Aut(X))=M_{F/K}(P:X\to X/Aut(X))$

And as consequence we have el teorema de Dèbes and Ensalem [P. Dèbes, M. Emsalem, On fields of Moduli. J. of Algebra. 1999]

Theorem: (Dèbes-Emsalem)
Let $\small K$ < $\small F$ be a finite Galois extension, $\small X$ a non- singular projective algebraic curve of genus $\small g\geq 2$ defined over $\small F$ , and $\small K=M_{F/K}(X)$ . Let $\small Aut(X)$ be the birational automorphism group of $\small X$ defined over $\small F$ . Let $\small P:X\to X/Aut(X)$ be the Galois covering with cover group $\small Aut(X)$ . Then:
1.- Exist a non-singular projective algebraic curve $\small B\simeq X/Aut(X)$ , defines over $\small K$ (called canonic model of $\small X/Aut(X)$ ), and an isomorphism $\small R:X/Aut(X)\to B$ defined over $\small F$ , such that if $\small Q=R\circ P$ , then $\small M_{F/K}(Q:X\to B)=K$ .
2.- Moreover, if we can fiend a point in $\small B-B_Q$ , where $\small B_Q$ is the set of all critical values of $\small Q$ , which are $\small K$ -rational, then $\small K$ is a definition field of $\small X$ .

PROOF: of Debes-Emsalem’s Theorem

More fields of Moduli

Another way to see fields of Moduli is through morphisms.

Let $K$ < $F$ be an extension fields, let $X,Y$ be projective varieties defined over $F$ and let $P:X\to Y$ be an algebraic morphism of finite degree over $F$ . We also assume that $Y$ is defined over $M_{F/K}(X)$ .

Applying $\sigma \in Aut(F/K)$ to $P:X\to Y$ , we obtain a new morphism $P^\sigma:X^\sigma\to Y^\sigma$ . Since $Y$ is defined over $M_{F/K}(X),$ $Y=Y^\sigma$ , this implies that $P^\sigma:X^\sigma\to Y$ .

It said that the morphism $P:X\to Y$ is equivalent to $P^\sigma:X^\sigma\to Y$ , which is denoted $\{P:X\to Y\}\simeq\{P^\sigma:X^\sigma\to Y\}$ , if there exists a birational isomorphism $f_\sigma:X\to X^\sigma$ , such that the following diagram commutes
$\xymatrix{ X \ar[r]^{f_\sigma} \ar[d]_{P} & X^\sigma \ar[ld]^{P^\sigma}\\ Y & }$
that is, $P^\sigma\circ f_\sigma=P$ .

Definition: (Field of Moduli for morphism)
Let $K$ < $F$ be an extension field, let $X,Y$ be projective algebraic varieties defined over $F$ and let $P:X\to Y$ be an algebraic morphism of finite degree over $F$ . We also assume that $Y$ is defined over $M_{F/K}(X)$ . Consider the following subgroup of $Aut(F/K)$

$F_K(P:X\to Y)=\{\sigma\in Aut(F/K):\{P:X\to Y\}\simeq\{P^\sigma:X^\sigma\to Y\}\}$
We define the field of Moduli $M_{F_K}(P:X\to Y)$ of $P:X\to Y$ , respect to the extension fields $K$ < $F$ , as the fixed field of the subgroup $F_K(P:X\to Y)$ , i.e.

$M_{F_K}(P:X\to Y)=Fix(F_K(P:X\to Y))$

Remark: Notice form the definition that $M_{F/K}(X)< M_{F/K}(P:X\to Y)$ , but in general the might be different.

If the morphism $P:X\to Y$ is a Galois morphism, that is $P$ is defined by the action of a finite subgroup $G$ of $Aut(X)$ , then for each $\sigma\in Aut(F/K)$ we can see the group $G^\sigma=\{\gamma^\sigma:\gamma\in G\}$ < $Aut(X^\sigma)$ .

Note that, if $\sigma\in F_K(P:X\to Y)$ , than it is not totally clear the existence of an isomorphism $f_\sigma:X\to X^\sigma$ , such that $f_\sigma G f\sigma^{-1}=G^\sigma$ . But, if the group $G$ is unique (in some sense), for example when $G= Aut(X)$ , then it is true and we have the next result.

Theorem: Let $X$ be a non-singular projective algebraic variety, $G$ < $Aut(X)$ a finite group of birational automorphism of $X$ and $P:X\to Y$ an algebraic morphism by the action of $G$ so that $Y$ is defined over $M_{F/K}(X)$ . If the group $G$ is unique in $Aut(X)$ , then

$\begin{array}{rcl} F_K&=&F_K(P:X\to Y)\\[0.3cm] M_{F/K}&=&M_{F/K}(P:X\to Y) \end{array}$

First post

As a first post, I’m going to give some definitions and theorems that I’ve learned in these two years.

We are going to start with a projective variety $X\subset \mathbb{P}^n(F)$ . If $f$ $\in$ $\mathbb{C}$ $[x_0,...,x_n]$ is an homogeneous polynomial of degree $n$ , then we have the associated projective algebraic curve

$X(f)=\{[x,y,z]\in\mathbb{P}^2(\mathbb{C}):f(x,y,z=0)\}$

If $f$ in non- singular (i.e their partial derivates do not vanish simultaneously at any point of $X(f)$ ) then we can endow to $X(f)$ with a compact Riemann surface structure.

So, here is our first definition

Definition: (Fields of definition)

Let $K$ < $F$ be an extension field and $X$ $\subset$ $\mathbb{P}^n(F)$ let be a projective algebraic variety. A field of definition for $X$ is any field $N$ , such that $K$ < $N$ < $F$ , so that there exists a projective algebraic variety $Y$ birational equivalent to $X$ ( $Y\cong X$ ).

Now, consider $K$ < $F$ an extension field, Aut( $F/K$ ) its automorphism group and $F$ $[x_0,...,x_n]$ the polynomial ring of $F$ with unknown variables $x_0,...,x_n$ .

If $f=\sum a_{i_0},...,a_{n}x_0^{i_0}\cdots x_n^{i_n}\in F[x_0,...,x_n]$

and , $\sigma\in$ Aut( $F/K$* )* then we have the natural action of $\sigma$ on $f$ given as

$\sigma(f)=f^\sigma=\sum\sigma(a_{i_0},...,i_n)x_0^{i_0}\cdots x^{i_n}$

This provides a natural action of Aut( $F/K$ ) over the ring $F$ $[x_0,...,x_n]$ .

If we consider $X$ $\subset$ $\mathbb{P}^n(F)$ a projective algebraic variety defined by

$X=\{f_1=...=f_r\}$

then every $\sigma\in$ Aut( $F/K$ ) provides a new projective algebraic variety $X^\sigma$ defined by

$X^\sigma=\{f_1^\sigma=...=f_r^\sigma\}$

Theorem: Let $K$ < $F$ be an extension field and let $X\subset$ $\mathbb{P}^n(F)$ , $Y\subset \mathbb{P}^m(F)$ be birational equivalent projective algebraic varieties. Then, for every $\sigma\in$ Aut( $F/K$ ) it holds that $X^\sigma$ and $Y^\sigma$ are also birational equivalent.

Given $X$ , a natural question is how different can be $X$ and $X^\sigma$ , when $\sigma$ run through Aut( $F/K$ ).
If we denote by $\mathcal{C}$ the set of birational equivalent classes of projective algebraic varieties defined over $F$ , then we have that the action described above induces in a natural way the following action
$\begin{array}{ccccc} \bullet&:&Aut(F/K)\times\mathcal{C}&\to&\mathcal{C}\\ &&(\sigma,[X])&\mapsto&[X^\sigma] \end{array}$
The orbit of [ $X$ ] is conformed by all the classes of algebraic varieties [ $Y$ ] which are birational equivalent to $X^\sigma$ , for all $\sigma$ $\in$ Aut( $F/K$ ), that is,
$Orb([X])=\{[Y]\in\mathcal{C}:[Y]=[X^\sigma],\ \forall \sigma\in Aut(F/K)\}$
The stabilizer of [ $X$ ], with respect to the above action, is given by
$\begin{array}{rcl} Stab([X])=F_K(X)&=&\{\sigma\in Aut(F/K):[X^\sigma][X]\}\\[0.3cm] &=&\{\sigma\in Aut(F/K):X\cong X^\sigma\} \end{array}$

A field of Moduli is the smallest field where a Riemann surface can be defined, its definition is

Definition: (Field of Moduli)

Let $K$ < $F$ be an extension field an let $X\subset$ $\mathbb{P}^n(F)$ be a projective algebraic variety. The field of Moduli for $X$ , with respect to the extension field $K$ < $F$ , is
$M_{F/K}=Fix(F_K(X))=\{a\in F: \sigma(a)=a,\ \forall\sigma\in Aut(F/K))\}$

Remark:

Let $K$ < $F$ be an extension fields and let $X\subset$ $\mathbb{P}^n(F)$ be a projective algebraic variety.

1.- By the above definition, it easy to see that $K$ < $M_{F/K}(X)$ < $F$ .
2.- If $K$ < $F$ is a general Galois extension and $X$ is defined over $K$ , then $M_{F/K}(X)$ = $K$ .

Indeed: Since $X$ is defined over $K$ , then for every $\sigma\in$ Gal( $F/K$ ), $X^\sigma = X$ .

Thereby,

$F_K(X)=Gal(F/K) \mbox{ and } M_{F/K}(X)=K$
3.- Let $N$ be a field such that, $K$ < $N$ < $F$ . Since Aut( $F/N$ ) is a subgroup of Aut( $F/K$ ), we have that $M_{F/K}(X)$ is a subfield of $M_{F/N}(X)$ . But in general,

$M_{F/N}(X)\ne M_{F/K}(X)$

Now we are going to see how field of Moduli are related ti the fields of definition.

Theorem: Let $K$ < $F$ be a general Galois extension and let $X\subset$ $\mathbb{P}^n(F)$ be a projective algebraic variety. Then, every field of definition of $X$ contains the field of Moduli of $X$ .

PROOF: Let $K$ < $N$ < $F$ be an extension fields, where $N$ is a fields of definition for $X$ . We can assume that $X$ is defined by homogeneous polynomials with coefficients in $N$ .
If we take $\sigma\in$ Aut( $F/N$ ) < Aut( $F/K$ ), then $X^\sigma=X$ , this means $\sigma\in$ $F_K(X)$ , which implies Aut( $F/N$ ) < $F_K(X)$ . Therefore,

$M_{F/K}(X)=Fix(F_K(X))< Fix(Aut(F/N))=N\quad _\square$

Another theorem that could be useful
Theorem: Let $F$ be an algebraically closed field, let $K$ be the prime field of $F$ and let $X\subset$ $\mathbb{P}^n(F)$ be a projective algebraic variety. Then, there exists a field of definition for $X$ which is a finite extension of the field of Moduli $M_{F/K}(X)$ .
PROOF: See H. Hammer – F. Herrlich, “A remark on the Moduli Field of a Curve” Arch. Math. 81. 2003.

Note that the above theorem works for projective algebraic varieties defined over $F=\mathbb{C}$ and taking $K=\mathbb{Q}$ .

To end this first post, I will state a theorem whose proof is not hard, but you must be careful with writing.

Theorem: Let $K$ < $F$ be an extension field, where $F$ is algebraically closed, and let $X\subset$ $\mathbb{P}^n(F)$ be a projective algebraic variety. Let $\overline{M_{F/K}(X)}$ be the algebraic closure of the field of Moduli $M_{F/K}(X)$ in $F$ . Then,

$M_{F/K}(X)=M_{\overline{M_{F/K}(X))}/M_{F/K}(X)}(X)$

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