Primitive algebraic points on curves

A number field $K$ is primitive if $K$ and $\mathbb{Q}$ are the only subextensions of $K$. Let $C$ be a curve defined over $\mathbb{Q}$. We call an algebraic point $P\in C(\overline{\mathbb{Q}})$ primitive if the number field $\mathbb{Q}(P)$ is primitive. We present several sets of sufficient conditions for a curve $C$ to have finitely many primitive points of a given degree $d$. For example, let $C/\mathbb{Q}$ be a hyperelliptic curve of genus $g$, and let $3 \le d \le g-1$. Suppose that the Jacobian $J$ of $C$ is simple. We show that $C$ has only finitely many primitive degree $d$ points, and in particular it has only finitely many degree $d$ points with Galois group $S_d$ or $A_d$. However, for any even $d \ge 4$, a hyperelliptic curve $C/\mathbb{Q}$ has infinitely many imprimitive degree $d$ points whose Galois group is a subgroup of $S_2 \wr S_{d/2}$.


Introduction
By a curve C over a field K we mean a smooth projective and geometrically irreducible variety defined over K having dimension 1.We say that a curve C defined over a field K is hyperelliptic if the genus of C is at least 2 and C admits a degree 2 morphism C → P 1 , defined over K, which we shall refer to as the hyperelliptic morphism.We say that C/K has K-gonality m if m is the least degree of a non-constant morphism π : C → P 1 defined over K. Thus, for example, a hyperelliptic curve defined over K has K-gonality 2. We say that C/K is bielliptic if its genus is at least 2 and it admits a degree 2 morphism C → E, defined over K, where E is a curve of genus 1.
Recall that a number field K is called imprimitive if there is some subextension Q L K; if there is no such subextension then K is called primitive.Now let C be a curve defined over Q.Let P ∈ C be an algebraic point; i.e.P ∈ C(Q).We say P has degree d if the number field Q(P ) has degree d.We say that P is primitive if the number field Q(P ) is primitive, otherwise we say that P is imprimitive.A degree d point P ∈ C(Q) is called P 1 -isolated if there is no degree d non-constant morphism φ : C → P 1 , defined over Q, such that P is in the preimage of an element of P 1 (Q).The notion of isolated points was introduced in [7] and has become important in understanding low degree points on curves, particularly on modular curves (e.g.[8], [18]).It is easy to see that if d < m, where m is the Q-gonality of C, then P is P 1 -isolated.A key observation we make in this paper is that primitive points are often P 1 -isolated even if the degree is greater than the gonality.Theorem 1.Let C be a curve defined over Q with genus g and Q-gonality m ≥ 2. Let d ≥ 2 be an integer satisfying (1) d = m, d < 1 + g m − 1 .
Let P ∈ C(Q) be a degree d point on C that is not P 1 -isolated.Then Q(P ) contains a subfield of index d ′ satisfying In particular, the following hold.
(I) If gcd(d, m) = 1 or d is prime then any degree Under certain additional assumptions on the Jacobian J of the curve C, it is possible to conclude finiteness of primitive degree d points on C. Theorem 2. Let C be a curve defined over Q with genus g and Q-gonality m ≥ 2. Let d ≥ 2 be an integer satisfying (1).Let J be the Jacobian of C. Suppose either of the following hold: (a) J(Q) is finite; (b) or d ≤ g − 1, and A(Q) is finite for every abelian subvariety A/Q of J of dimension ≤ d/2.Then C has finitely many primitive degree d points.Moreover, if gcd(d, m) = 1 or d is prime then C has finitely many degree d points.
Observe that, for d ≤ g − 1, assumption (b) is trivially satisfied if J is simple.We moreover note that the inequality d ≤ g − 1 in (b) follows immediately from (1) if m ≥ 3.
Example 3. We consider the modular curve X 0 (239) which has genus g = 20 and Q-gonality m = 6 (see [32,Table 3]).We consider degree d points for d = 2, 3, 4; we note that (1) is satisfied for these values of d.A straightforward computation in Magma, which makes use of modular symbols algorithms due to Cremona [12] and Stein [40], shows that the Jacobian J 0 (239) of X 0 (239) factors as where A 3 and A 17 are simple abelian varieties of dimension 3 and 17, respectively.Moreover A 17 has analytic rank 0, and A 3 has positive analytic rank.Assuming the Birch and Swinnerton-Dyer conjecture, J 0 (239)(Q) is infinite, and so hypothesis (a) of Theorem 2 is not satisfied.However, J 0 (239), clearly satisfies hypothesis (b) for d = 2, 3, 4. By Theorem 2, we conclude that X 0 (239) has finitely many quadratic, cubic and primitive quartic points.
We point out the following theorem [17,Proposition 2.3] which gives a stronger conclusion than Theorem 2 under different assumptions.
Theorem 4 (Derickx and Sutherland).Let C be a curve defined over Q with genus g and Q-gonality m ≥ 2. Let J be the Jacobian of C and suppose J(Q) is finite.Suppose d < m.Then C has finitely many degree d points.
The following corollary to Theorem 2 is obtained by restricting Theorem 2 and its proof to the hyperelliptic case.
Corollary 5. Let C be a hyperelliptic curve defined over Q with genus g.Let J be the Jacobian of C and let d be a positive integer.Suppose either of the following hold: Then C has finitely many primitive degree d points.More precisely, the following hold.
(i) If d is odd, then C has finitely many degree d points.
(ii) If d is even, then for all but finitely many degree d points P on C, the field Q(P ) contains a subfield of index 2.
We note that Gunther and Morrow [21, Proposition 2.6] prove that conclusions (i) and (ii) hold for 100% of genus g hyperelliptic curves over Q with a rational Weierstrass point, provided only that d ≤ g − 1, though their proof is rather different.
In contrast to Corollary 5 we note the following.
Lemma 6.Let C be a hyperelliptic curve defined over Q.Let d ≥ 4 be an even integer.Then C has infinitely many imprimitive degree d points.
Proof.We may suppose C has an affine model where F ∈ Q[X] is a squarefree polynomial.Let L be any number field of degree d/2 and choose θ ∈ L such that L = Q(θ).By Faltings' theorem, C(L) is finite.Thus we may choose some a ∈ Q such that F (θ + a) is a non-square in L. Let P = (θ + a, F (θ + a)).This is a degree d point on C, and is imprimitive as Q(P ) contains the index 2 subfield L.
Let C be a curve defined over Q.Let P ∈ C be an algebraic point; i.e.P ∈ C(Q).We define the Galois group of P to be the Galois group of the Galois closure of Q(P )/Q.A degree d ≥ 4 point whose Galois group is S d or A d is primitive (Lemma 29).Thus if d is an even integer and if C, d satisfy the hypotheses of Corollary 5 then C has only finitely many degree d points with Galois group S d or A d .However, it follows from the proof of Lemma 6 that C has infinitely many degree d points with Galois group contained in S 2 ≀ S d/2 .In a separate paper [27] we explore Galois groups of algebraic points in more detail.For now, we content ourselves with the following result.
Theorem 7. Let C be a hyperelliptic curve defined over Q with genus 2 or 3. Let J be the Jacobian of C and suppose that J(Q) is trivial.Then C has no quartic points with Galois group S 4 or A 4 .However, C has infinitely many quartic points with Galois group contained in D 4 .
The above results exploit the gonality map C → P 1 to make deductions about low degree primitive points.However, the existence of a low degree map C → C ′ with C ′ of positive genus also makes it more likely for low degree primitive algebraic points on C to be P 1 -isolated, as illustrated by the following theorem.Theorem 8. Let π : C → C ′ be a morphism of curves defined over Q of degree m ≥ 2. Write g, g ′ for the genera of C, C ′ respectively, and suppose g ′ ≥ 1.Let d ≥ 2 be an integer satisfying In particular, the following hold.
Theorem 9. Let π : C → C ′ be a morphism of curves defined over Q of degree m ≥ 2. Write g, g ′ for the genera of C, C ′ respectively, and suppose g ′ ≥ 1.Let d ≥ 2 be an integer satisfying (3).Write J for the Jacobian of C and suppose J(Q) is finite.Then C has finitely many primitive degree d points.Moreover, if gcd(d, m) = 1 or d is prime then C has finitely many degree d points.
Applying Theorem 9 and its proof to bielliptic curves gives the following.
Corollary 11.Let C be a bielliptic curve defined over Q with genus g.Let J be the Jacobian of C and suppose Then C has finitely many primitive degree d points.More precisely, the following hold.
(i) If d = 2 or d is odd, then C has finitely many degree d points.
(ii) If d ≥ 4 and even, then for all but finitely many degree d points P on C, the field Q(P ) contains a subfield of index 2.
The above theorems are concerned with finiteness criteria for low degree primitive points.In the opposite direction, we show that if the degree d is sufficiently large compared to the genus, then there are infinitely many primitive degree d points, provided there is at least one such point.
Theorem 12. Let C/Q be a curve.Let d ≥ g + 1 where g is the genus of C. Suppose there exists a primitive degree d point on C. Then there are infinitely many primitive degree d points on C.
The paper is organized as follows.In Section 2 we review some standard results regarding Riemann-Roch spaces and we use these to prove Theorem 7. In Section 3 we recall the Castelnuovo-Severi theorem and use it to give proofs of Theorems 1 and 8.In Section 4 we give criteria for when a complete linear series does not contain any primitive divisors (a primitive divisor is simply the Galois orbit of a primitive algebraic point).Let C satisfy the hypotheses of either Theorem 2 or Theorem 9, and write C (d) for the d-th symmetric power of C. In Section 5, using a powerful theorem of Faltings, we show that C (d) (Q) may be decomposed as a finite union of complete linear series.In Section 6 we prove Theorems 2 and 9 and their corollaries, making use of the results developed in previous sections.In Section 7 we recall the relationship between primitive extensions and primitive group actions, and we use this together with Hilbert's irreducibility theorem to give a proof of Theorem 12. Finally, in Sections 8 and 9 we give consequences of Theorems 2 and 9 (and their proofs) for the modular curves X 1 (N ) and X 0 (N ) for certain small N .
We are grateful to Nils Bruin, Victor Flynn, Samuel Le Fourn, David Loeffler, Filip Najman, Petar Orlić, Damiano Testa and Bianca Viray for helpful discussions, and to the referees for suggesting many improvements.

Proof of Theorem 7
Let C be a curve defined over Q.When we speak of divisors on C we in fact mean rational divisors: a divisor on C/Q is a finite formal integral linear combination D = a i P i of algebraic points P i that is stable under the action of Gal(Q/Q).We call this divisor effective and write D ≥ 0 if and only if a i ≥ 0 for all i.An irreducible divisor is an effective divisor that cannot be written as the sum of two non-zero effective divisors.Thus an effective degree d divisor D is irreducible if and only if there is a degree d point P ∈ C(Q) such that D = P 1 + P 2 + • • • + P d where {P 1 , . . ., P d } is the Galois orbit of P .We say that D is the irreducible divisor corresponding to P .
For a divisor D on C we denote by L(D) the corresponding Riemann-Roch space defined by and we let ℓ(D) = dim L(D).We shall make frequent use of the Riemann-Roch theorem [3, page 13] which asserts that ( 4) here K C is any canonical divisor on C, and g is the genus of C. Recall that deg and cannot be linearly equivalent to an effective divisor.In that case ℓ(K C −D) = 0. We shall also require Clifford's theorem [22,Theorem IV.5.4] on special effective divisors.Recall that an effective divisor D is special if ℓ(K C − D) > 0.
Theorem 13 (Clifford).Let D be an effective special divisor on a curve C. Then Moreover, equality occurs if and only if D = 0, or D is a canonical divisor, or C is hyperelliptic and D is a multiple of a hyperelliptic divisor.
Recall that a hyperelliptic curve C is equipped with a degree 2 morphism π : Proof of Theorem 7. Let C be as in the statement of Theorem 7. We may suppose C has an affine model as in (2) where F ∈ Q[X] is a squarefree polynomial of degree 2g + 1 or 2g + 2. It follows from Lemma 6 and its proof that C has infinitely many quartic points with Galois group contained in D 4 .To complete the proof it is enough to show that there are no quartic points on C with Galois group S 4 or A 4 .
Let P be a degree 4 point on C, and let D be the corresponding irreducible divisor.Since where f ∈ L(D 0 ).We claim that 1, X, X 2 is a Q-basis of L(D 0 ).Let us first assume our claim and use it to complete the proof.Thus ) is a quartic field, and Y (P ) 2 = F (X(P )), we see that Q(X(P )) is quadratic and contained in the quartic field Q(P ).Therefore the Galois group of P is neither S 4 nor A 4 .It remains to prove our claim.Note that X has a double pole at infinity and no other poles if deg(F ) = 2g + 1; and also X has a simple pole at ∞ + and ∞ − , and has no other poles if deg(F ) = 2g + 2. Therefore, 1, X, X 2 belong to L(D 0 ), and so ℓ(D 0 ) ≥ 3. It is enough to show that ℓ(D 0 ) = 3.We now make use of our assumption that g = 2 or 3.If g = 2 then Riemann-Roch immediately gives ℓ(D 0 ) = 3. Suppose g = 3.Then Riemann-Roch tells us that D 0 is special, and since it is twice a hyperelliptic divisor, Clifford's theorem gives the equality ℓ(D 0 ) = 3.

Proofs of Theorems 1 and 8
We start by recalling the classical Castelnuovo-Severi theorem.
Theorem 14 (Castelnuovo-Severi theorem).Let k be a perfect field, and let X, Y , Z be curves over k.Denote the genera of these curves by g(X), g(Y ) and g(Z) respectively.Let π Y : X → Y and π Z : X → Z be non-constant morphisms defined over k, having degrees m and n respectively.Suppose Then there is a curve X ′ defined over k, and a morphism X → X ′ also defined over k and of degree > 1 through which both π Y and π Z factor.
Proof.We are unable to find a reference that gives the precise statement that we need.Indeed the theorem is most often given in the context of complex Riemann surfaces (for example [1]), or the field of definition of the morphism X → X ′ is not mentioned (for example [24,Corollary]).We therefore give an explanation of how the version above follows straightforwardly from a proof due to Mattuck and Tate [28] The theorem follows from our claim on letting X ′ be the normalization of D.
We prove our claim by contradiction.Suppose deg(h) = 1.The h is a birational map, and g(D) = g(X) (where g(D) denotes the geometric genus of D).Mattuck and Tate [28, page 296] show that ( 6) where K S denotes any canonical divisor on S (the inequality on the left is itself referred to as the Castelnuovo-Severi inequality).By the adjunction formula for surfaces [22 where p a (D) is the arithmetic genus of D. Thus, from (6), However, since the geometric genus is bounded by the arithmetic genus, contradicting assumption (5).This establishes our claim.
Proof of Theorem 1.Let P be a degree d point on C and suppose P is not P 1isolated.Thus there is a degree d map f : C → P 1 , defined over Q, such that f (P ) = α ∈ P 1 (Q).Observe that f * (α) consists precisely of the Galois orbit of P .Let m ≥ 2 be the Q-gonality of C; thus in particular, there is a morphism π : C → P 1 defined over Q, of minimal degree m ≥ 2. As m and d satisfy (1), the Castelnuovo-Severi theorem immediately implies that the morphisms f , π must simultaneously factor through a common non-trivial morphism of curves.We obtain a commutative diagram of non-constant morphisms of curves ( 7) Thus the field Q(P ) of degree d contains the subfield Q(Q) of index d ′ .The theorem follows.
Proof of Theorem 8.This proof is similar to the proof of Theorem 1.As before, let P be a degree d point on C and suppose that P is not P 1 -isolated.Thus there is a degree d map f : From assumption (3), the Castelnuovo-Severi theorem gives a commutative diagram of non-constant morphisms of curves ( 8) The rest of the proof is similar to the proof of Theorem 1.

Primitive Divisors in Complete Linear Series
Let C be a curve over Q.For an effective divisor D on C the notation |D| denotes the complete linear series containing D: this is precisely the set of effective divisors linearly equivalent to D. Observe that for effective divisors D, D ′ , we have The following well-known lemma highlights the relationship between a point being not P 1 -isolated and the complete linear series of its irreducible divisor having positive dimension.
Lemma 15.Let C be a curve defined over Q, and let d ≥ 1.Let D be an irreducible degree d divisor.The following are equivalent.
It is clear that (c) implies (d).To complete the proof we suppose (d) and prove (c).Thus there is a morphism f : C → P 1 of degree d defined over Q with f (P ) = α ∈ P 1 (Q).Composing f with an automorphism of P 1 we may suppose α = ∞.Now P ∈ div ∞ (f ), and so D ≤ div ∞ (f ) as div ∞ (f ) is stable under the action of Galois.Thus D = div ∞ (f ) as both divisors have degree d.This completes the proof.
We will call an irreducible divisor primitive if it is the Galois orbit of a primitive point.

Decomposition of C (d) into complete linear series
We denote the d-th symmetric power of C by C (d) .Recall that C (d) (Q) can be identified with the set of effective degree d divisors on C. The purpose of this section is to prove the following proposition.
To prove the proposition we shall need the following famous theorem of Faltings [19].
Theorem 19 (Faltings).Let B be an abelian variety defined over a number field K, and let V ⊂ B be a subvariety defined over K. Then there is a finite number of abelian subvarieties B 1 , . . ., B m of B, defined over K, and a finite number of points x 1 , . . ., x m ∈ V (K) such that the translates x i + B i are contained in V , and, moreover, such that We shall also need the following theorem of Debarre and Fahlaoui [13, Corolllary 3.6].
Theorem 20 (Debarre and Fahlaoui).Let C/C be a curve of genus g ≥ 1 with Jacobian J, and let d ≤ g − 1.Let D 0 be a fixed divisor of degree d on C. Let W d (C) be the image of C (d) under the Abel-Jacobi map (11) ι : Let A be an abelian subvariety of J with a translate contained in Proof of Proposition 18.If C (d) (Q) = ∅ then we take n = 0 and there is nothing to prove.So suppose d) under the Abel-Jacobi map (11), which is defined over Q.We claim that By the theorem of Debarre and Fahlaoui, dim Observe that if C and d satisfy the hypotheses of either Theorem 2 or 9 then they satisfy the hypotheses of Proposition 18 and therefore C (d) (Q) can be decomposed into the union of finitely many complete linear series as in (9).Suppose d is even.By Proposition 18, we have that (9) holds where D 1 , . . ., D n is a finite set of effective degree d divisors on C. Let P be a degree d point and let D be the corresponding irreducible divisor.Then D ∈ |D i | for some i.Suppose D = D i .Then ℓ(D) ≥ 2 and so by Lemma 15 the point P is not P 1 -isolated.It follows from Theorem 1 that Q(P ) contains a subfield of index d ′ = 2. Thus, for all but finitely many degree d points P we have that Q(P ) contains a subfield of index 2.
Proof of Theorem 9.This follows from Proposition 18 and Corollary 17 by trivial modifications to the proof of Theorem 2.
Proof of Corollary 11.The proof is a straightforward modification of preceding arguments.
6.1.A Remark on Effectivity.Let C and d satisfy the hypotheses of Theorem 2 or Theorem 9. Then C (d) (Q) can be decomposed into a finite union of complete linear systems as in (9).Suppose that we are able to explicitly compute the representatives D i in (9).Then we have an effective strategy for computing all primitive degree d points.Indeed, if ℓ(D i ) ≥ 2 then |D i | contains no primitive divisors by Corollary 16.We are left to consider |D i | for ℓ(D i ) = 1.However, if ℓ(D i ) = 1, then |D i | = {D i } and we simply need to test D i to determine if it is the Galois orbit of a primitive degree d point.Moreover, if gcd(d, m) = 1 or d is prime then we can compute all degree d points by a slight modification of the strategy: if ℓ(D i ) = 1 then simply test D i for irreducibility.
We remark that the decomposition (9) can often be computed using symmetric power Chabauty (e.g.[39] or [10]) provided r + d ≤ g where r is the rank of the Mordell-Weil group J(Q).

6.2.
A Remark on the Assumptions of Corollary 5.As noted previously, assumption (b) of Corollary 5 is satisfied for d ≤ g − 1 whenever the Jacobian J is simple.We remark that almost all hyperelliptic curves have simple Jacobians, in a sense that we make precise shortly.For this we will need the following theorem of Zarhin [44, Theorem 1.1].
Theorem 21 (Zarhin).Let k be a field, char(k) = 2. Let C : Y 2 = f (X), where f ∈ k[X] is a separable polynomial of degree n ≥ 5, and let J denote the Jacobian of C. Suppose char(k) = 3 or n ≥ 7, and that f has Galois group either S n or A n .Then End(J) ∼ = Z, and in particular J is absolutely simple.
We fix a genus g ≥ 2. Let n = 2g+1 or 2g+2.The set of all polynomials of degree ≤ n can be naturally identified with A n+1 (Q): here a = (a 0 , a 1 , . . ., a n ) ∈ A n+1 (Q) corresponds to the polynomial Hilbert's irreducibility theorem asserts the existence of a thin set S ⊂ A n+1 (Q) such that for a ∈ A n+1 (Q) − S, the polynomial f a is irreducible of degree n and has Galois group S n .See [38,Chapter 9] for a statement of Hilbert's irreducibility theorem as well as the definition of thin sets.
Therefore, for a ∈ A n+1 (Q) − S, the genus g hyperelliptic curve Y 2 = f a (X) has a simple Jacobian by Zarhin's theorem.
We point out that there is no shortage of hyperelliptic curves satisfying the finite Mordell-Weil group condition of Corollary 5.This immediately follows from the preceding remarks together with the following Theorem of Yu [43,Theorem 3], applied with r = 0.
Theorem 22 (Yu).Let K be a number field with at least one real embedding.Let f ∈ K[X] be a separable degree n polynomial such that n ≡ 3 (mod 4) and Gal(f ) ∼ = S n or A n .Let C : Y 2 = f (X) be a hyperelliptic curve over K with Jacobian J. Then for every r ≥ 0, there are infinitely many quadratic twists J b of J such that dim F2 (Sel 2 (J b /K)) = r.

Proof of Theorem 12
The proof of Theorem 12 relies on the following proposition, which is in fact a consequence of Hilbert's Irreducibility Theorem.
Proposition 23.Let d ≥ 2 be an integer.Let f : C → P 1 be a degree d morphism of curves defined over Q, and let α ∈ P 1 (Q).Suppose that α is not a branch value for f , and that the fibre f −1 (α) consists of a single Galois orbit of points.Let P ∈ f −1 (α) and suppose the extension Q(P ) is primitive.Then there is a thin set S ⊂ P 1 (Q) such that for β ∈ P 1 (Q) − S, the fibre f −1 (β) consists of a single Galois orbit of points and, for any Before proving Proposition 23 we show that it implies Theorem 12.
Proof of Theorem 12. Let C/Q be a curve of genus g.Suppose P C(Q) is primitive of degree d ≥ g + 1, and let D be the corresponding irreducible divisor.
By Riemann-Roch, It follows from Lemma 15 that there is a degree d morphism f : We apply Proposition 23 with α = ∞ ∈ P 1 (Q).The theorem follows.
The proof of Proposition 23 makes use of the relationship between primitive extensions and primitive Galois groups.Whilst this relationship is known, we are unable to find a convenient reference, and we therefore give the details.Let X be a non-empty set, and let G be a group acting transitively on X.The trivial partitions of X are {X} and {{x} : x ∈ X}.A partition P of X is said to be G-stable if σ(Z) ∈ P for all σ ∈ G and Z ∈ P. Observe, as the action of G on X is transitive, that G also acts transitively on any G-stable partition P, and that any two elements of P therefore have the same cardinality.
We say that G acts imprimitively on X if X admits a G-stable non-trivial partition.If X does not have a G-stable non-trivial partition then we say that G acts primitively on X.The following lemma is an immediate consequence of this definition.
Lemma 24.Let G be a group acting transitively on a set X. Let G ′ be a subgroup of G and suppose that G ′ acts primitively on X.Then G acts primitively on X.
The following result is well-known, and in particular implies that S d and A d act primitively on {1, . . ., d}, for d ≥ 1 and d ≥ 3 respectively.Lemma 25.Let G be a group acting 2-transitively on a set X. Then the action is primitive.
Proof.Let P be a G-stable partition of X and suppose Y ∈ P has at least two elements.We would like to show that Y = X.Lemma 26.Let G be a group acting transitively on a set X.The action is imprimitive if and only if there is a proper subset Y of X, with at least two elements, such that Proof.Given a G-stable non-trivial partition P we can take Y to be any element of P. As P is a partition, Y clearly satisfies (12), and as P is non-trivial, Y is a proper subset of X with at least two elements.Conversely, suppose Y is a proper subset of X containing at least two elements and satisfying (12).We easily check that P = {τ (Y ) : τ ∈ G} is a G-stable non-trivial partition.
Lemma 27.Let G be a finite group acting transitively on a non-empty finite set X. Let x ∈ X, and write Stab(x) for the stabilizer of x in G.The action of G on X is imprimitive if and only if Stab(x) is a non-maximal proper subgroup of G.
Proof.Let x ∈ X and assume the existence of a subgroup Stab(x) 12) is satisfied and so the action is imprimitive.
Conversely, suppose the existence of a proper subset Y of X with at least two elements satisfying (12).As the action is transitive, we may in fact suppose that Then there is some σ ∈ G such that σ(x) = x ′ .Thus σ(Y ) = Y , and so σ ∈ H but σ / ∈ Stab(x).It follows that Stab(x) is a proper in H, and so is non-maximal as a subgroup of G.
Lemma 28.Let K = Q(θ) be a number field and let K be its Galois closure over Q.Let G = Gal( K/Q).Let d = [K : Q] and let θ 1 , . . ., θ d ∈ K be the Galois conjugates of θ.Then G acts primitively on {θ 1 , . . ., θ d } if and only if the extension K/Q is primitive.
Proof.Let X = {θ 1 , . . ., θ d }.Then G acts transitively on X.We let x = θ ∈ X and note that the stabilizer Stab(θ) is in fact Gal( K/K).By the Galois correspondence, K is imprimitive if and only if the subgroup Gal( K/K) is proper and non-maximal in G, which by Lemma 27 if equivalent to the action of G being imprimitive.
The following result was assumed in the introduction.Lemma 29.Let C/Q be a curve, let d ≥ 3 and let P be a degree d point on C with Galois group S d or A d .Then P is primitive.
Proof.The lemma follows from Lemma 28, and for this we need the fact that S d and A d act primitively on {1, . . ., d}.This is trivially true if d = 3, and for d ≥ 4 follows from Lemma 25.
Proof of Proposition 23.By composing f with a suitable automorphism of P 1 we may suppose that α = 0 ∈ P 1 (Q).Write K = Q(C) for the function field of C. We may regard f as a non-constant element of K, and with this interpretation By hypothesis, α = 0 is not a branch value for f , and f −1 (0) consists of a single Galois orbit containing P .Write D = f * (0) which we think of as a degree d place of K, unramified in the extension K/Q(P 1 ).Write , for the valuation ring of D and its maximal ideal.Then the residue field O D /m D can be identified with K = Q(P ) where the identification is given by g + m D → g(P ).Now fix θ ∈ K such that K = Q(θ).Then there is some g ∈ O D such that g(P ) = θ.As g ∈ K and K has degree of degree n | d, such that gcd(a 0 (V ), . . ., a n (V )) = 1, and F (g, f ) = 0. Now, F (θ, 0) = F (g(P ), f (P )) = 0, and so θ is a root of the polynomial this polynomial is non-zero as gcd(a 0 (V ), . . ., a n (V )) = 1.As θ has degree d, it follows that n = d, and that F (U, V ) is irreducible over Q(V ).In particular F (U, V ) = 0 is a (possibly singular) plane model for C. As C is absolutely irreducible, F (U, V ) is irreducible over Q. Write K for the Galois closure of K/Q(P 1 ), and let g 1 , . . ., g d be the roots of Let C/Q be the algebraic curve associated to K. Let D be an extension of the place D of K to K. As D is unramified in K/Q(P 1 ), the place D is unramified in the Galois closure K/Q(P 1 ) (see for example [41,Corollary III.8.4]).We claim that g 1 , . . ., g d ∈ O D. To see this, note that G = Gal(K/Q(P 1 )) acts transitively on the g i .Thus we would like to show ord D(τ (g)) ≥ 0 for all τ ∈ G.However, where the second equality follows as g ∈ K and τ −1 ( D) is an unramified place of K above D.This proves the claim.
The place D corresponds to a Galois orbit of points on C and we let P be a representative point chosen above P .Write G D for the decomposition group corresponding to D in G = Gal( K/Q(P 1 )); by definition this is As D is unramified in the Galois closure K/Q(P 1 ), the theory of decomposition groups allows us to identify the decomposition group G D with Gal( K/Q), where K is the Galois closure of K/Q.Here we shall in fact need the details of the precise identification.Let σ ∈ G D .We associate σ to σ ′ ∈ Gal( K/Q) as follows.We let γ ∈ K ∼ = O D/m D. Then γ = h( P ) for some h ∈ O D , and we define σ ′ (γ) = σ(h)( P ).It immediately follows from the definition of G D that σ ′ is welldefined.The map G D → Gal( K/Q) given by σ → σ ′ in fact an isomorphism [41,Theorem III.8.2], and from now on we shall identify Gal( K/Q) with G D via this identification.
Let θ 1 , . . ., θ d be the conjugates of θ in K, where we take θ 1 = θ.Now θ = g(P ) = g( P ).Let 1 ≤ i ≤ d.Then there is some σ i ∈ G D such that σ i (θ) = θ i and therefore σ i (g)( P ) = θ i .Thus there is a conjugate of g sending P to θ i .As the θ i are pairwise distinct, we may after reordering the g i suppose that g i ( P ) = θ i .Now Gal( K/Q) = G D is a subgroup of G = Gal( K/Q(P 1 )) and both act on {g 1 , . . ., g d }.
As K/Q is primitive, Lemma 28 tells us that G D acts primitively on the θ i and therefore on the g i .Thus by Lemma 24, the action of G on {g 1 , . . ., g d } is primitive.
Finally we apply Hilbert's Irreducibility Theorem [38,Chapter 9] to f : C → P 1 .For β ∈ P 1 (Q) we shall make a standard abuse of language and speak of the decomposition group G β by which we mean the decomposition group G Q for any point Q ∈ C above β.As usual, G β is only defined up to conjugation in G. Now, Hilbert's Irreducibility Theorem applied to f : C → P 1 asserts the existence of a thin set S ⊂ P 1 (Q), which includes all branch values, such that G β = G for β ∈ P 1 (Q) − S. We enlarge S by adjoining finitely many values in P 1 (Q): the roots of a d (V ) (which is the leading coefficient of F regarded as an element of Q[V ][U ]); the V -coordinate of any singular point of the plane model F (U, V ) = 0; and the point ∞ ∈ P 1 (Q).As we have added finitely many points to the set S it remains thin.Let β ∈ P 1 (Q) − S. Let φ 1 , . . ., φ d be the roots of F (U, β).Let Q = (φ 1 , β); this is a smooth point on the plane model, and so may be regarded as a point on C. Let Q be a point of C above Q.Then as before, we can identify the action of G β on {φ 1 , . . ., φ d } with the action of G on {g 1 , . . ., g d }.As G is acting transitively and primitively on {g 1 , . . ., g d }, we have that G β = Gal(Q(f −1 (β))/Q) is acting transitively and primitively on {φ 1 , . . ., φ d }.Hence, the Galois action on the fibre f −1 (β) = {(φ 1 , β), . . ., (φ d , β)} is primitive and, by Lemma 28, the field ) is primitive of degree d.The proposition follows.
Let C/Q be a curve and let d ≥ g + 1 where g is the genus of C. Theorem 12 asserts the existence of infinitely many primitive degree d points on C provided there is at least one.However, the existence of a primitive degree d point is not guaranteed, as illustrated by the following lemma.
Lemma 30.Let g ≥ 2 be even.Let C be a degree 2g + 1 genus g curve defined over Q Suppose J(Q) = 0 where J is the Jacobian of C. Then C has no points of degree g + 1 points.
Proof.Write ∞ for the single point at infinity on the given model.Write D 0 = (g + 1)∞.Note that X has a double pole at ∞. Thus 1, X, . . ., X g/2 ∈ L(D 0 ).We claim that 1, X, . . ., X g/2 is a basis for L(D 0 ).We first explain how our claim implies the lemma.Let D be an effective degree g + 1 divisor.Since J(Q) = 0, we have for some α 0 , . . ., α g/2 ∈ Q.In particular f ∈ L(g∞).Thus is effective.Hence D is reducible.It follows that C has no degree g + 1 points.
In a positive direction, we can use Theorem 12 to construct curves with infinitely many primitive points.
Then there is a hyperelliptic curve C/Q of genus g with infinitely many primitive degree d points.
Proof.Let K = Q(θ) be any primitive number field of degree d.Let θ 1 , . . ., θ d be the conjugates of θ in a fixed Galois closure K of K. Choose a rational number α such that 2α = θ i + θ j for any pair 1 ≤ i, j ≤ d.Let φ = θ − α.The conjugates of φ are φ i = θ i − α with 1 ≤ i ≤ d, and satisfy φ i = ±φ j for any pair i, j.Note that be the minimal polynomial of φ 2 , which must be irreducible of degree d.Let h = f (X 2 ).The roots of h are ±φ 1 , . . ., ±φ d which are pairwise distinct and hence h is separable of degree 2d = 2g + 2. Let C be the genus g hyperelliptic curve Note that this has the primitive degree d point (φ, 0).Hence by Theorem 12 there are infinitely many primitive degree d points.
8. Low degree primitive points on some X 1 (N ) Mazur [29] showed that all rational points on X 1 (p) are cuspidal for prime p ≥ 11.Merel's uniform boundedness theorem [30] asserts that for prime p, and for d satisfying (3 d/2 + 1) 2 ≤ p, the only degree d points on X 1 (p) are cuspidal.We now know, for each 1 ≤ d ≤ 8, the exact set of primes p such that X 1 (p) has degree d non-cuspidal points, thanks to the efforts of Kamienny [23], Parent [36], [37], Derickx, Kamienny, Stein and Stoll [15], and Khawaja [26].Less is known about the low degree points on X 1 (N ) for composite N , though several authors consider the somewhat easier problem of determining the values of N such X 1 (N ) has infinitely many degree d points for given small d (see for example [7] and [17] for two different approaches to studying this problem).Example 10 illustrates how our results can be applied to modular curves X 1 (N ) provided the analytic rank of J 1 (N ) is 0 and we have information about the quotients or gonality of X 1 (N ).The LMFDB [42] contains a database of modular curves X 1 (N ) for 1 ≤ N ≤ 70.For 61 of these curves the Jacobian J = J 1 (N ) has analytic rank 0. It follows from a theorem of Kato [25,Corollary 14.3] that the Mordell-Weil group J(Q) is finite.We are able to apply Theorem 9 to around half of these curves in order to deduce the finiteness of primitive points of certain low degrees.We note that it is common for X 1 (N ) to cover multiple curves, and in these instances we apply Theorem 9 to the covered curve C ′ that gives the most generous range for d in inequality (3).We record the results in Table 1.
In [14,Theorem 3.1] the authors give a complete list of N for which J 1 (N ) has analytic rank 0. There are in total 83 such values of N , the largest of which is N = 180.9. Low degree primitive points on some X 0 (N ) The computational study of quadratic points on modular curves is an active area of research (see e.g.[2], [11], [20], [33], [35], to name but a few works).
Comparatively less is known about points defined over number fields of higher degree.Still, there is reason to be hopeful.Establishing the modularity of all elliptic curves over totally real cubic fields [16], and totally real quartic fields not containing √ 5 [9] required the study of cubic, and quartic points on certain modular curves.Banwait and Derickx [4] have determined all cubic points on X 0 (N ) for N ∈ {41, 47, 59, 71}.Box, Gajović, and Goodman [10] have determined all cubic points on X 0 (N ) for N ∈ {53, 57, 61, 65, 67, 73}, and all quartic points on X 0 (65).A famous theorem of Ogg [34] asserts that there are 19 values of N for which which X 0 (N ) is hyperelliptic.Of these, the only one for which J 0 (N )(Q) is infinite is N = 37 For these N , the quadratic points on X 0 (N ) have been determined by Bruin and Najman [11].It is easy to apply Corollary 5 to these curves and derive conclusions about algebraic points of degree 3 ≤ d ≤ g, where g is the genus of X 0 (N ).For example, consider C = X 0 (71) with genus g = 6.By Corollary 5 we know that there are only finitely many points on X 0 (71) of degrees 3 and 5, and finitely many primitive points of degrees 4, 6.We point out that we can in fact go further and compute these finite sets of points, as sketched in Remark 6.1.We illustrate this by giving some details of the computation of primitive degree 6 points on X 0 (71), making use of information found in [11] concerning the model and the Mordell-Weil group.A model for X 0 (71) is given by X 0 (71) : Y 2 = X 14 + 4X The only rational points are the two rational points at infinity which we denote by ∞ + and ∞ − (these are in fact the two cusps of X 0 (71)).Write where J = J 0 (71).Let P be a primitive degree 6 point on X 0 (71), and let D be the corresponding effective irreducible degree 6 divisor.Hence [D − 3D ∞ ] ∈ J(Q).
It follows that  2. This table gives the conclusions of our computations of primitive points on X 0 (N ) of certain low degrees d and for the values of N is the first column.Here g is the genus of X 0 (N ), and J(Q) is in fact the structure of the Mordell-Weil group where J = J 0 (N ).The table gives the number of primitive degree d points on X 0 (N ) up to Galois conjugacy.The symbol − indicates that our method is inapplicable for that particular N and d.  3.For each pair (N, d), the table gives a description of the effective degree d divisors D with ℓ(D) = 1 on the modular curve X 0 (N ).We denote by n d,r the number of such divisors that are reducible, n d,p the number of such divisors that are irreducible and primitive, and n d,i the number of such divisors that are irreducible but imprimitive.The symbol − indicates that we did not carry out the computation for the pair (N, d).
of the Castelnuovo-Severi inequality.Let S = Y × Z.Given two divisors D, D ′ on the surface S, we denote their intersection number by D • D ′ .Let D be the image of X on S under the morphism (π Y , π Z ) : X → S, and write h : X → D for the induced map.Clearly the curve D and the map h are defined over k.Let u : D → Y and v : D → Z be the restrictions of the projections Y

Corollary 16 .
Let C be a curve defined over Q with genus g and Q-gonality m ≥ 2. Let d ≥ 2 be an integer satisfying(1).Let D ′ be an effective degree d divisor onC with ℓ(D ′ ) ≥ 2. Then |D ′ | contains no primitive degree d divisors.Moreover, if gcd(d, m) = 1 or d is prime, then |D ′ | contains no irreducible divisors.Proof.Suppose D ∈ |D ′ | is an irreducible divisor.Then ℓ(D) = ℓ(D ′ ) ≥ 2.By Lemma 15, any P ∈ D is not P 1 -isolated.By part (II) of Theorem 1 we see that P is imprimitive.Thus |D ′ | contains no primitive divisors.Suppose now that gcd(d, m) = 1 or d is prime.Then part (I) of Theorem 1 gives a contradiction, therefore |D ′ | contains no irreducible divisors.The following variant of Corollary 16 has an almost identical proof, but appealing to Theorem 8 instead of Theorem 1. Corollary 17.Let π : C → C ′ be a morphism of curves defined over Q of degree m ≥ 2. Write g, g ′ for the genera of C, C ′ respectively, and suppose g ′ ≥ 1.Let d ≥ 2 be an integer satisfying (3).Let D ′ be an effective degree d divisor on C with ℓ(D ′ ) ≥ 2. Then |D ′ | contains no primitive degree d divisors.Moreover, if gcd(d, m) = 1 or d is prime, then |D ′ | contains no irreducible divisors.

Proposition 18 .
Let C be a curve over Q of genus g ≥ 1, and let J be the Jacobian of C. Let d be a positive integer.Suppose either of the following two conditions hold:(a) J(Q) is finite; (b) or d ≤ g − 1, and A(Q) is finite for every abelian subvariety A/Q of J of dimension ≤ d/2.Then there are finitely many effective degree d divisors D 1 , D 2 , . . ., D n such that(9)

6 .
Proofs of Theorems 2 and 9 and their corollaries Proof of Theorem 2. Let C, m, d be as in the statement of Theorem 2. By Proposition 18, C (d) (Q) has a finite decomposition, as in (9) where D 1 , . . ., D n are effective degree d divisors.If ℓ(D i ) ≥ 2 then, by Corollary 16, the complete linear series |D i | does not contain primitive divisors.On the other hand, if ℓ(D i ) = 1 then |D i | = {D i }.Hence there are only finitely many primitive degree d points on C. Suppose now that gcd(d, m) = 1 or d is prime.Again, if ℓ(D i ) ≥ 2 then, by Corollary 16, the complete linear series |D i | does not contain irreducible divisors.The theorem follows.Proof of Corollary 5. Let C/Q be hyperelliptic of genus g.This has gonality m = 2. Suppose either of hypotheses (a), (b) of Corollary 5 is satisfied.Then C, g, m, d satisfy the hypotheses of Theorem 2. In particular, if d is odd then C has finitely many degree d points.
Let a, b ∈ Y be distinct, and let c ∈ X be distinct from a, b.Then there is some σ ∈ G such that σ(a) = a and σ(b) = c.Thus a ∈ Y ∩ σ(Y ) which forces σ(Y ) = Y , and therefore c ∈ Y .Hence Y = X.