Aven Bross

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Belyi's Theorem

Dec 2, 2018

Belyi’s Theorem is an arithmetic-geometric result that has the unusual quality of being both fairly recent and provable with only the material from a first course in classical algebraic geometry.

In what follows $\mathbb{A}^n,\mathbb{P}^n$ are over $\mathbb{C}$ unless otherwise specified.

Theorem (Belyi, 1979). Let $C\subset\mathbb{P}^2$ be an irreducible non-singular projective curve defined as the zero set of $F\in\overline{\mathbb{Q}}[X,Y,Z].$ Then there exists a finite regular map $C\to\mathbb{P}^1$ ramified only over a subset of ${0,1,\infty}.$

The proof strategy takes a projection $\pi$ of $C$ onto $\mathbb{P}^1$ and then use a sequence of regular maps ${\psi_i}$ to gather the ramification points of $\pi$ into the set ${0,1,\infty}.$

Lemma 1. The projection away from the point $(0:0:1)$ is a finite regular map $\pi:C\to\mathbb{P}^1$ ramified over the points $\pi(P)$ such that $F_Z(P)=0.$

Let $Q=(x_0:y_0)\in\mathbb{P}^1.$ Then $\pi$ is ramified over $Q$ if and only if there exists $P\in\pi^{-1}(Q)$ such that $e_P(\pi)>1.$ Note that

is a local parameter for $Q$, where $H(X,Y)$ is any linear form not vanishing at $Q.$ Thus

Hence $e_P(\pi)>1$ if and only if $y_0X-x_0Y$ is zero on the tangent line $\Theta_{C,P}.$ The tangent line is described by

We care whether the tangent line contains the line $y_0X-x_0Y$, i.e., if the tangent line passes through $(0:0:1).$ This is equivalent to $F_Z(P)=0.$

Claim 1. The map $\pi$ is ramified only over points in $\overline{Q}\cup{\infty}.$

By Lemma 1 above we know that the ramification points of $\pi$ are the images of points in $C\cap V(F_Z).$ By Bezout’s Theorem

contains precisely $\text{deg}F\cdot\text{deg}F_Z$ points when counted with multiplicity. By Bezout again, $C\cap V(F_Z)$ contains precisely the same number of points when considered in $\mathbb{P}^2(\mathbb{C}).$ Since $\mathbb{P}^2(\overline{\mathbb{Q}})\subset\mathbb{P}^2(\mathbb{C})$ all intersection points are contained in $\mathbb{P}^2(\overline{\mathbb{Q}}).$

Lemma 2. A regular map $\psi:\mathbb{A}^1\to\mathbb{A}^1$ defined by $x\mapsto G(x)$ is ramified precisely over the images of points $x_0$ where $G_X(x_0)=0.$ Hence it extends to a map $\mathbb{P}^1\to\mathbb{P}^1$ ramified over only over those affine ramification points and $\infty$.

Note that if $\psi$ is ramified over $y_0\in\mathbb{A}^1$ if and only if there exists $x_0\in\mathbb{A}^1$ such that $G(x_0)=y_0$ and $e_{x_0}(\psi)>1.$

Observe that

We find that

and thus $G_X=H_X.$ Thus if $G_X(x_0)=0$, then

and $(X-x_0)^2\mid H(X).$ Conversely if we write $H(X)=(X-x_0)^mu(X)$ for some unit $u(X)\in\mathcal{O}_{\mathbb{A}^1,x_0}$ then $H_X(x_0)=0$ if and only if $m>1.$

Let $\varphi:C\to\mathbb{P}^1$ be a regular map ramified only over a finite set $S\subset\overline{\mathbb{Q}}\cup{\infty}.$

Claim 2. If $S\subset\overline{\mathbb{Q}}\cup{\infty}$, then there exists $\psi:\mathbb{P}^1\to\mathbb{P}^1$ such that $\psi\circ\phi$ is ramified only over points in $\mathbb{Q}\cup{\infty}.$

Let $n$ be the maximum degree over $\mathbb{Q}$ of an element in $S$ and $p$ be the number of elements in $S$ of degree $n.$ We will proceed by double induction on $n$ and $p.$

Our object is to construct a map $\psi:\mathbb{P}^1\to\mathbb{P}^1$ such that $\psi\circ\phi$ is ramified only over points in $S’\subset \overline{\mathbb{Q}}\cup{\infty}$ where the number of elements in $S’$ of degree $n$ over $\mathbb{Q}$ is less than $p$ and the maximum degree of elements in $S’$ is less than or equal to $n.$

Let $\alpha\in S$ be an element of degree $n$ over $\mathbb{Q}$ and let $G\in\mathbb{Q}[X]$ be its minimal polynomial. Define $\psi:\mathbb{P}^1\to\mathbb{P}^1$ by $x\mapsto G(x).$ Let ${\beta_1,\ldots,\beta_r}$ be the roots of $G_X.$ Note that since $\text{deg}(G_X)<n$ the elements $\beta_1,\ldots,\beta_r$ are of degree $n-1$ or fewer over $\mathbb{Q}.$

Observe that ramification points of $\psi\circ\phi$ are

Moreover, applying a polynomial over $\mathbb{Q}$ cannot increase the degree of an algebraic element over $\mathbb{Q}.$ Thus the maximum degree over $\mathbb{Q}$ of elements in $S’$ is $n$ and there are at most $p-1$ such elements since $\phi(\alpha)=0.$

Claim 3. If $S\subset\mathbb{Q}\cup{\infty}$ then there exists a regular map $\psi:\mathbb{P}^1\to\mathbb{P}^1$ such that $\psi\circ\varphi$ is ramified only over points in ${0,1,\infty}.$

If $|S|\le 3$ then clearly there is an automorphism $\psi$ such that $\psi(S)\subset{0,1,\infty}.$

Suppose that $|S|\ge 4.$ We will construct a map $\psi:\mathbb{P}^1\to\mathbb{P}^1$ such that $\psi\circ\varphi$ is ramified over some set of points $S’\subset\mathbb{Q}\cup{\infty}$ with $|S’|<|S|.$

By applying an appropriate automorphism of $\mathbb{P}^1$ we may assume that $S={0,1,\infty,\lambda_1,\ldots,\lambda_r}.$ Subsequently we may apply the automorphisms $(X,Y)\mapsto (1-X,Y)$ and $(X,Y)\mapsto (Y,X)$ as necessary to ensure that $0<\lambda_1<1$; note that each maps ${0,1,\infty}\to{0,1,\infty}.$

Write $\lambda_1=m/(m+n).$ Let $c=\frac{(m+n)^{m+n}}{m^nn^m}.$ Let us define $\psi:\mathbb{P}^1\to\mathbb{P}^1$ by $x\mapsto G(x)$ where $g(X)=cX^m(1-X)^n.$ Observe that

which has zeros only at $0,1$ and $\lambda_1.$ Thus $\psi$ is ramified only over

Continuing inductively and composing we construct a map $\psi:\mathbb{P}^1\to\mathbb{P}^1$ such that $\psi(S)\subset{0,1,\infty}$ and $\psi$ is ramified only over the values in $S.$

Belyi’s Theorem follows from claims 1, 2, and 3.

Example. Consider the elliptic curve $C$ defined by the affine equation

Note the curve has one point “at infinity,” namely the point $(0:1:0).$ Let $\pi$ be the projection away from this point, that is, $(x,y)\mapsto x$ on the affine patch $z=1.$ Note that that this extends to a projection $C\to\mathbb{P}^1$ since $C$ is nonsingular. This is a slightly different projection than the one in the proof above, but works nicely in this case.

Observe that $\text{deg}(\pi)=2$ since $[\mathbb{C}(C):\mathbb{C}(X)]=2.$ Therefore projection $\pi$ is ramified over $\infty$ and the images of the points in $C$ where $F_Y(X,Y)=2Y=0.$ These are the points $0,1,\sqrt{2}.$

The minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $X^2-2.$ Thus we consider the map $\psi_1$ defined by $t\mapsto t^2-2$; note this map is ramified only at $0.$ We compute

We next apply the map $\phi_2$ defined by $t\mapsto -1/t$ which achieves the dual purpose of mapping ${0,-1,\infty}$ to ${0,1,\infty}$ and placing $\psi_2(-2)=1/2\in (0,1).$ Finally, we apply the Belyi map $t\mapsto 4t(1-t)$ which sends ${0,1,\infty,1/2}$ to ${0,1,\infty}.$