Let C be an algebraic curve defined on number field K . We note C ( K ) be the set of algebraic points on C defined on K and ∪ [ K : ℚ ] ≤ l     C ( K ) the set of algebraic points on C to be coordinated in K of degree at most l over ℚ . The degree of an algebraic point R is the degree of its defining field on ℚ ; deg ( R ) = [ ℚ ( R ) : ℚ ] . A famous theorem of Faltings states that if g ≥ 2 then the set C ( K ) of algebraic points on C defined on K is finite. A generalization to subvarieties of an abelian variety allows a qualitative study of the set ∪ [ K : ℚ ] ≤ l     C ( K ) of algebraic points on C of degree at most l over ℚ .

We propose to study in detail the set of algebraic points of any degree given on ℚ on the curve C 3 ( 11 ) of affine equation y 11 = x 3 ( x − 1 ) 3 .

Our affine equation curve C 3 ( 11 ) : y 11 = x 3 ( x − 1 ) 3 is a special case of quotients of Fermat curves of equations C r , s ( p ) : y p = x r ( x − 1 ) s , 1 ≤ r , s , r + s ≤ p − 1 studied in [1].

Let P 0 = ( 0 : 0 : 0 ) , P 1 = ( 1 : 0 : 1 ) and P ∞ = ( 1 : 0 : 0 ) denote the point at infinity of C 3 ( 11 ) . Consider the Jacobian folding defined by

j : C 3 ( 11 ) ( ℚ ) → J ( ℚ ) P ↦ [ P − P ∞ ]

We will designate J the Jacobian of C 3 ( 11 ) and by j ( P ) the class denoted [ P − P ∞ ] of P − P ∞ .

Our approach relies on the knowledge of the Mordell-Weil group of the Jacobian J-variety of C 3 ( 11 ) and the condition that it is finite: it consists in using the Abel-Jacobi theorem to plunge the curve into its Jacobian and to study linear systems on the curve C 3 ( 11 ) .

The Mordell-Weil group J ( ℚ ) of rational points of the Jacobian J of C 3 ( 11 ) is finite and given by J ( ℚ ) ≅ ( ℤ / 11 ℤ ) ( [2], p. 219 and [3]).

Our study results from the work of Gross-Rohrlich who determined ∪ [ K : ℚ ] ≤ 2     C 1 ( 11 ) ( K ) the set of algebraic points on C 1 ( 11 ) ( K ) of degree at most 2 on ℚ and given by the following proposition:

Proposition 1.

The set of algebraic points on C 1 ( 11 ) ( K ) of degree at most 2 on ℚ is given by

∪ [ K : ℚ ] ≤ 2     C 1 ( 11 ) ( K ) = { ( 1 2 ± y 11 + 1 4 , y ) } ∪ { P ∞ } (1)

We extend these results by giving a geometric parametrization of algebraic points of any given degree on ℚ on the curve C 3 ( 11 ) of affine equation y 11 = x 3 ( x − 1 ) 3 .

Our essential tools are:

1) The Mordell-Weil group J ( ℚ ) of the Jacobian of C .

2) The Abel-Jacobi theorem (see in [4] page 156).

3) The study of linear systems on the curve C 3 ( 11 ) .

4) The theory of intersection.

Our main result is as follows:

Theorem

The set of algebraic points of degree l ≥ 9 on C 3 ( 11 ) is:

∪ [ K : ℚ ] ≤ l     C 3 ( 11 ) ( K ) = F 0 ∪ ( ∪ k = 1 10     F k ) (2)

with

F 0 = { ( − ∑ i ≤ l 2       a i y i 3 ∑ j ≤ l − 11 2     b j y j 3 , y ) | a 0 ≠ 0 , a l 2 ≠ 0   if   l   is   even ,                   b l − 11 2 ≠ 0   if   l   is   odd   and   y   root   of   the   equation               y 11 ( ∑ j ≤ l − 11 2     b j y j ) 2 = ( ∑ i ≤ l 2     a i y i ) ( ∑ i ≤ l 2     a i y i 3 + ∑ j ≤ l − 11 2     b j y j 3 ) 3 } (3)

F k = { ( − ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 ∑ j ≤ l − k 2     b j y j 3 , y ) | b 0 ≠ 0 , a l + 11 − k 2 ≠ 0   if   l   is   even ,                 b l − k 2 ≠ 0   if   l   is   odd   and   y   root   of   the   equation             y k ( ∑ j ≤ l − k 2     b j y j ) 2 = ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i − ( 11 − k ) ) ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 + ∑ j ≤ l − k 2     b j y j 3 ) 3 } (4)

Let x and y be the rational functions defined on C 3 ( 11 ) by: x ( X , Y , Z ) = X Z and y ( X , Y , Z ) = Y Z .

For a divisor D on C 3 ( 11 ) , let L ( D ) be the ℚ ¯ -vector space of the rational functions f defined by

L ( D ) = { f ∈ ℚ ¯ ( C 3 ( 11 ) ) * | d i v ( f ) ≥ − D } ∪ { 0 } (5)

The projective equation of the curve C 3 ( 11 ) is: Y 11 = X 3 Z 5 ( X − Z ) 3 .

We have the following Lemma:

Lemma 1

C 3 ( 11 ) : y 11 = x 3 ( x − 1 ) 3

d i v ( x ) = 11 P 0 − 11 P ∞ ;

d i v ( y ) = 3 P 0 + 3 P 1 − 6 P ∞ ;

d i v ( x − 1 ) = 11 P 1 − 11 P ∞ .

Proof 1 It is a calculation of type

d i v ( x − i ) = ( ( X − i Z ) = 0 ) . C 3 ( 11 ) − ( Z = 0 ) . C 3 ( 11 ) (6)

From (6), we have d i v ( x ) = ( X = 0 ) . C − ( Z = 0 ) . C .

For X = 0 , the projective equation gives Y 11 = 0 ; and for Z = 1 , we obtain the point P 0 = ( 0 : 0 : 1 ) of multiplicity equal to 11.

For Z = 0 , the projective equation gives Y 11 = 0 ; and for X = 1 , we obtain the point P ∞ = ( 1 : 0 : 0 ) of multiplicity equal to 11. Thus d i v ( x ) = 11 P 0 − 11 P ∞ .

In the same way we show that d i v ( x − 1 ) = 11 P 1 − 11 P ∞ and d i v ( y ) = 3 P 0 + 3 P 1 − 6 P ∞ .

Consequence 1

11 j ( P 0 ) = 11 j ( P 1 ) = 0 ;

3 j ( P 0 ) + 3 j ( P 1 ) = 0

so j ( P 0 ) and j ( P 1 ) generate the same subgroup J ( ℚ ) .

Lemma 2 A ℚ -base of L ( l P ∞ ) is given by :

B = { ( x 2 ( x − 1 ) 2 y 7 ) i | i ∈ ℕ , i ≤ l 2 } ∪ { x ( x 2 ( x − 1 ) 2 y 7 ) j | j ∈ ℕ , j ≤ l − 11 2 } (7)

Proof 2. It is clear that B is free. It remains to show that

d i m ( B ) = d i m ( L ( l P ∞ ) ) .

By the Riemann-Roch theorem, we have d i m ( L ( l P ∞ ) ) = l − g + 1 as soon as l ≥ 2 g − 1 with g = 11 − 1 2

Let us consider the following cases:

Case 1: Suppose that l is even, and let l = 2 h . Then we have

i ≤ l 2 = h

and

j ≤ l − 11 2 ⇔ j ≤ 2 h − 11 2 ⇔ j ≤ 2 h − 11 − 1 2 = h − 6 = h − g − 1 .

So we obtain

B = { 1 , x 2 ( x − 1 ) 2 y 7 , ⋯ , ( x 2 ( x − 1 ) 2 y 7 ) h } ∪ { x , x x 2 ( x − 1 ) 2 y 7 , ⋯ , x ( x 2 ( x − 1 ) 2 y 7 ) h − g − 1 } ,

and therefore d i m ( B ) = ( h + 1 ) + ( h − g ) = 2 h − g + 1 = l − g + 1 = d i m ( L ( l P ∞ ) ) .

Case 2: Suppose that l is odd, and let l = 2 h + 1 .

i ≤ l 2 ⇔ i ≤ 2 h + 1 2 ⇔ i ≤ h

and

j ≤ l − 11 2 ⇔ j ≤ 2 h − 10 2 = h − g

So we obtain

B = { 1, x 2 ( x − 1 ) 2 y 7 , ⋯ , ( x 2 ( x − 1 ) 2 y 7 ) h } ∪ { x , x x 2 ( x − 1 ) 2 y 7 , ⋯ , x ( x 2 ( x − 1 ) 2 y 7 ) h − g } ,

and therefore

d i m ( B ) = ( h + 1 ) + ( h − g + 1 ) = 2 h + 1 − g + 1 = l − g + 1 = d i m ( L ( l P ∞ ) ) .

Let R ∈ C 3 ( 11 ) ( ℚ ¯ ) with [ ℚ ( R ) : ℚ ] = l . Let R 1 , ⋯ , R l be the Galois conjugates of R, and let t = [ R 1 + ⋯ + R l − l P ∞ ] which is a point of J ( ℚ ) = { m j ( P 0 ) ,0 ≤ m ≤ 10 } ; so t = m j ( P 0 ) with 0 ≤ m ≤ 10 . This gives the relation

[ R 1 + ⋯ + R l − l P ∞ ] = m j ( P 0 ) . (8)

We note that R ∉ { P 0 , P 1 , P ∞ } .

Case m = 0

Then there exists a rational function f such that d i v ( f ) = R 1 + ⋯ + R l − l P ∞ , so f ∈ L ( l P ∞ ) . According to Lemma 2, we have

f = ∑ i ≤ l 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i + x ∑ j ≤ l − 11 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j

with a l 2 ≠ 0 if l is even (otherwise the R i would be equal to P ∞ ) and b l − 11 2 ≠ 0 if l is odd (otherwise the R i would be equal to P ∞ ). At the points R i we have

∑ i ≤ l 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i + x ∑ j ≤ l − 11 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j = 0

hense

x = − ∑ i ≤ l 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i ∑ j ≤ l − 11 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j

and therefore

y 11 = x 3 ( x − 1 ) 3 ⇔ y 1 3 = x 2 ( x − 1 ) 2 y 7 ,

so

x = − ∑ i ≤ l 2     a i y i 3 ∑ j ≤ l − 11 2     b j y j 3 .

So the equation y 11 = x 3 ( x − 1 ) 3 becomes

y 11 ( ∑ j ≤ l − 11 2     b j y j ) 2 = ( ∑ i ≤ l 2     a i y i ) ( ∑ i ≤ l 2     a i y i 3 + ∑ j ≤ l − 11 2     b j y j 3 ) 3

which is an equation of degree l in y. We thus find a family of points of degree l

F 0 = { ( − ∑ i ≤ l 2     a i y i 3 ∑ j ≤ l − 11 2     b j y j 3 , y ) | a 0 ≠ 0 , a l 2 ≠ 0   if   l   is   even ,                 b l − 11 2 ≠ 0   if   l   is   odd   and   y   root   of   the   equation             y 11 ( ∑ j ≤ l − 11 2     b j y j ) 2 = ( ∑ i ≤ l 2     a i y i ) ( ∑ i ≤ l 2     a i y i 3 + ∑ j ≤ l − 11 2     b j y j 3 ) 3 }

In the same way we show that for m = k with k ∈ { 1, ⋯ ,10 } , the relation (8) gives [ R 1 + ⋯ + R l − l P ∞ ] = k j ( P 0 ) = ( k − 11 ) j ( P 0 ) . Then there exists a rational function f such that d i v ( f ) = R 1 + ⋯ + R l + ( 11 − k ) P 0 − ( l + 11 − k ) P ∞ , so f ∈ L ( l + 11 − k ) P ∞ . According to the Lemma 2, we have f = ∑ i ≤ l + 11 − k 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i + x ∑ j ≤ l − k 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j ; and as o r d f P 0 = 11 − k ,

therefore

f = ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i + x ∑ j ≤ l − k 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j

with a l + 11 − k 2 ≠ 0 if l is even (otherwise the R i would be equal to P ∞ ) and b l − k 2 ≠ 0 if l is odd (otherwise the R i would be equal to P ∞ ). At the points R i we have

∑ 11 − k ≤ i ≤ l + 11 − k 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i + x ∑ j ≤ l − k 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j = 0

hense x = − ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i ( x 2 ( x − 1 ) 2 y 7 ) i ∑ j ≤ l − k 2     b j ( x 2 ( x − 1 ) 2 y 7 ) j and therefore x = − ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 ∑ j ≤ l − k 2     b j y j 3 .

So the equation y 11 = x 3 ( x − 1 ) 3 becomes

y k ( ∑ j ≤ l − k 2     b j y j ) 2 = ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i − ( 11 − k ) ) ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 + ∑ j ≤ l − k 2     b j y j 3 ) 3

which is an equation of degree l in y. We thus find a family of points of degree l

F k = { ( − ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 ∑ j ≤ l − k 2     b j y j 3 , y ) | b 0 ≠ 0 , a l + 11 − k 2 ≠ 0   if   l   is   even ,                 b l − k 2 ≠ 0   if   l   is   odd   and   y   root   of   the   equation               y k ( ∑ j ≤ l − k 2     b j y j ) 2 = ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i − ( 11 − k ) ) ( ∑ 11 − k ≤ i ≤ l + 11 − k 2     a i y i 3 + ∑ j ≤ l − k 2     b j y j 3 ) 3 }

The authors declare no conflicts of interest regarding the publication of this paper.

Balde, B.S., Diallo, M.M.D. and Sall, O. (2022) Algebraic Points of Any Degree l with ( l ≥ 9 ) over ℚ on the Affine Equation Curve C 3 ( 11 ) : y 11 = x 3 ( x − 1 ) 3 . Advances in Pure Mathematics, 12, 519-525. https://doi.org/10.4236/apm.2022.129039