Another proof of three points theorem

G.V.Belyi

April 25, 1997

In the paper “On Galois Extensions of a Maximal Cyclotomic Field” we proved the following theorem.

1. Theorem. A complete nonsingular algebraic curve X defined over a field of characteristic zero can be defined over \overline{ℚ} if and only if it can cover ℙ^{1} with three branching points 0 , 1 , ∞ .

See Izvestiya Ak.Nauk SSSR, ser. mat. 43(2) (1979) , 269-276. The aim of this notice is to give a new proof of this theorem. We adduce the both proofs because they are very short and explicit, in order to the reader can easily compare them.

Old Proof. The “if” part of the theorem follows from the Weil criterion, so we have to show the “only if” part. We do it in three steps.

Step 1: Suppose that X is defined over \overline{ℚ} . We choose an arbitrary nonconstant rational function t ∈ \overline{ℚ} ( X) . It defines a covering of ℙ^{1} with branching in some finite set of points B_{1} that are defined over \overline{ℚ} .

Step 2: Let h_{i} ( t) be the minimal polynomial over ℚ for the set of finite points in the set B_{i} and let

B_{i+1} = { h_{i} ( ξ) | h_{i}′ ( ξ) = 0 }

be the set of finite branching points for the polynomial h_{i} ( t) . It is clear that

deg h_{i+1} ≤ ( deg h_{i}) − 1 .

Consequently, in no more than deg h_{1} steps we must obtain a linear polynomial h_{l} . We now take

f ( t) = h_{l−1} ( h_{l−2} ( … ( h_{1} ( t)) …)) .

By construction the covering f ( t) : X → ℙ^{1} branches in some rational points set A.

Step 3: If we find a polynomial g_{A} ( f ) ∈ ℚ [ f ] that is only ramified over {0, 1, ∞ } , and such that g_{A} ( A) ⊆ {0, 1, ∞ } , then the composition g_{A} ( f ( t) ) will give us the required covering. We shall say that such a polynomial satisfies the condition ( b_{A}) .

If A = A_{1} ∪ { s } and g_{A_{1}} ( f ) has already been constructed we denote by A_{2} the set {0 , 1 , ∞ , g_{A_{1}} ( s)} . Now we can take g_{A} ( f ) to be the polynomial g_{A_{2}} ( g_{A_{1}} ( f ) ) . Hence it suffices to construct the polynomial g_{A_{3}} ( f ) for a set A_{3} consisting of three finite rational points and ∞ . Making a linear change of variables, we may assume that A_{3} = {0 , s , 1 , ∞ } , where 0 < s < 1 . If s = m/(m + n) , where m and n are natural numbers, then we set

g_{A_{3}} ( f ) = \frac{( m + n)^{m+n}}{m^{m} · n^{n}} · f^{m} · ( 1 − f )^{n} .

New Proof. It differs from the old one only in the third step. We take a rational function instead of the polynomial g_{A} ( f ) ∈ ℚ [ f ] . Let now

g_{A} ( f ) = ∏_{i=1}^{n} ( f − λ_{i})^{r_{i}} ∈ ℚ ( f ) , where ∑_{i=1}^{n} r_{i} = 0 ,

and A = { λ_{1} , …, λ_{n} , ∞ } . We may suppose without loss of generality that all these λ are integers, 0 = λ_{1} < λ_{2} < ⋯ < λ_{n} , and gcd ( λ_{2} , …, λ_{n}) = 1 . Note that the function g_{A} ( f ) ramifies only at zeros and poles of its logarithmic derivative

\frac{g′_{A} ( f )}{g_{A} ( f )} = ∑_{i=1}^{n} \frac{r_{i}}{f − λ_{i}} .

Denote by W = W ( λ_{1} , …, λ_{n})  the Vandermonde determinant;  and let
w_{i} = (−1)^{n−1} W ( λ_{1} , …, λ̂_{i} , …, λ_{n})

be the Vandermonde determinant, where the hat λ̂_{i} means that the corresponding λ_{i} is omitted. Besides, we denote by N = gcd ( w_{1} , …, w_{n}) . We see that the function

∑_{i=1}^{n} \frac{w_{i}}{f − λ_{i}} = W · ∏_{i=1}^{n} ( f − λ_{i})^{−1}

has the unique n-multiple zero at the infinity. Setting r_{i} = w_{i} / N we see that the rational function g_{A}(f) satisfies the condition ( b_{A}) . In more detail , the function g_{A}(f) takes λ_{1} , λ_{3} , … to 0 ; it sends λ_{2} , λ_{4} , … to ∞ . Besides , g_{A}(f) takes ∞ to 1 and does not ramifies outside A .

We obtain the new function g_{A}(f) in much more explicit way comparatively to the old way, where we used induction; and where the polynomial g_{A} depended on a choice of some ordering on the set A . Although, we lose the property of the function to be a polynomial; nevertheless, we won something; now we can control the bad reduction points for the covering g_{A} : ℙ^{1} → ℙ^{1} .

2. Claim. A rational prime number p is the bad reduction point for the covering g_{A} iff p divides the Vandermonde determinant W_{A} = W ( λ_{1} , …, λ_{n}) .

Proof. Evident.

We suppose n is even now and define the hyperelliptic curve Y_{A} by the equation

y^{2} = ( x − λ_{1} ) ( x − λ_{2} ) … ( x − λ_{n}) .

Let X be a Galois closure of the covering g_{A} , H_{A} = Gal ( X/ℙ^{1} ) , and h_{A} be the order of the group H . An interesting question arises: It seams should exist a connection between this order h_{A} and the hight of this hyperelliptic curve h ( Y_{A}) ?

3. Examples : Let A′ be the finite part of A . We list the all covering g_{A} for A′ ⊂ { ±1 , ±3 , ±5 } ≅ { 0 , 1 , 2 , 3 , 4 , 5 } with n = # ( A′) = 4 , 5 , 6 . Besides, we draw corresponding Grothendieck’s “dessins d’enfants” in the cases they are symmetric . We understand them as inverse images of the segment g_{A}^{−1} (−∞ , 0 ) .

Under the restrictions mentioned above we have five more cases: g is equal to

\frac{t (t − 3)^{8}}{(t − 4)^{3} (t − 2)^{6}} ,  \frac{(t − 5)^{2} t^{3} (t − 2)^{10}}{(t − 3)^{5} (t − 1)^{10}} ,  \frac{t (t − 4)^{15}}{(t − 5)^{6} (t − 3)^{10}} ,

\frac{t^{8} (t − 3)^{10}}{(t − 5)^{3} (t − 1)^{15}} ,  or  \frac{(t − 5) t^{4} (t − 2)^{20}}{(t − 3)^{10} (t − 1)^{15}} .

We leave it to the reader to make pictures. It is not hard to check that the Galois groups H corresponding to the coverings in this list either are the symmeric groups S_{n} or are the alternating groups A_{n} .

Finally, I should like to thank the MPI for hospitality.