Polynomial functions on number fields are nearly
injective
Michael Zieve
(joint work with Alex Carney and Ruthi Hortsch)
University of Michigan
May 26, 2012
Polynomials over the rational numbers
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by c 7→ f (c) is at most
6-to-1 outside a finite set.
This result is best possible:
The “finite set” cannot be avoided: there are polynomials inducing
any prescribed function on any finite set (Lagrange).
The “6” cannot be improved: for f (X ) := (X 3 − X )2 ,
t 2 − 2t 2t − 1 t2 − 1 f ± 2
=f ± 2
=f ± 2
t −t +1
t −t +1
t −t +1
for each t ∈ Q.
2 / 13
Polynomials over the rational numbers
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by c 7→ f (c) is at most
6-to-1 outside a finite set.
This result is best possible:
The “finite set” cannot be avoided: there are polynomials inducing
any prescribed function on any finite set (Lagrange).
The “6” cannot be improved: for f (X ) := (X 3 − X )2 ,
t 2 − 2t 2t − 1 t2 − 1 f ± 2
=f ± 2
=f ± 2
t −t +1
t −t +1
t −t +1
for each t ∈ Q.
2 / 13
Polynomials over the rational numbers
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by c 7→ f (c) is at most
6-to-1 outside a finite set.
This result is best possible:
The “finite set” cannot be avoided: there are polynomials inducing
any prescribed function on any finite set (Lagrange).
The “6” cannot be improved: for f (X ) := (X 3 − X )2 ,
t 2 − 2t 2t − 1 t2 − 1 f ± 2
=f ± 2
=f ± 2
t −t +1
t −t +1
t −t +1
for each t ∈ Q.
2 / 13
Number fields
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by α 7→ f (α) is at most
6-to-1 outside a finite set.
Theorem (Carney–Hortsch–Z)
For any number field K , any f ∈ K [X ] induces a map K → K which is at
most N-to-1 outside a finite set, where N is the largest integer such that
cos(2π/N) ∈ K .
Note: this is sharp for every K .
3 / 13
Number fields
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by α 7→ f (α) is at most
6-to-1 outside a finite set.
Theorem (Carney–Hortsch–Z)
For any number field K , any f ∈ K [X ] induces a map K → K which is at
most N-to-1 outside a finite set, where N is the largest integer such that
cos(2π/N) ∈ K .
Note: this is sharp for every K .
3 / 13
Number fields
Theorem (Carney–Hortsch–Z)
For any f ∈ Q[X ], the function Q → Q defined by α 7→ f (α) is at most
6-to-1 outside a finite set.
Theorem (Carney–Hortsch–Z)
For any number field K , any f ∈ K [X ] induces a map K → K which is at
most N-to-1 outside a finite set, where N is the largest integer such that
cos(2π/N) ∈ K .
Note: this is sharp for every K .
3 / 13
The exceptional polynomials
We can describe all f ∈ Q[X ] for which there is some number field K such
that f : K → K is (≥ 2)-to-1 over infinitely many elements of K .
We can describe all f ∈ Q[X ] for which f : Q → Q is (≥ 3)-to-1 over
infinitely many numbers. For instance, if f is (≥ 5)-to-1 infinitely often,
then f = g ◦ (X 3 − X )2 ◦ ` for some g , ` ∈ C[X ] with deg(`) = 1.
We cannot describe all f ∈ Q[X ] for which f : Q → Q is (≥ 2)-to-1 over
infinitely many points, and we do not believe that a reasonable description
is possible. The condition says that (f (X ) − f (Y ))/(X − Y ) = 0 has
infinitely many zeroes in Q × Q. When deg(f ) = 4, this asks which
members of a certain (infinite) collection of elliptic curves have positive
rank, a question much more difficult than the Birch–Swinnerton-Dyer
conjecture. (Bhargava will likely receive a Fields Medal for his proof that a
positive fraction of elliptic curves have rank zero; it is not believed that
there is a nice description of which curves these are.)
4 / 13
The exceptional polynomials
We can describe all f ∈ Q[X ] for which there is some number field K such
that f : K → K is (≥ 2)-to-1 over infinitely many elements of K .
We can describe all f ∈ Q[X ] for which f : Q → Q is (≥ 3)-to-1 over
infinitely many numbers. For instance, if f is (≥ 5)-to-1 infinitely often,
then f = g ◦ (X 3 − X )2 ◦ ` for some g , ` ∈ C[X ] with deg(`) = 1.
We cannot describe all f ∈ Q[X ] for which f : Q → Q is (≥ 2)-to-1 over
infinitely many points, and we do not believe that a reasonable description
is possible. The condition says that (f (X ) − f (Y ))/(X − Y ) = 0 has
infinitely many zeroes in Q × Q. When deg(f ) = 4, this asks which
members of a certain (infinite) collection of elliptic curves have positive
rank, a question much more difficult than the Birch–Swinnerton-Dyer
conjecture. (Bhargava will likely receive a Fields Medal for his proof that a
positive fraction of elliptic curves have rank zero; it is not believed that
there is a nice description of which curves these are.)
4 / 13
The exceptional polynomials
We can describe all f ∈ Q[X ] for which there is some number field K such
that f : K → K is (≥ 2)-to-1 over infinitely many elements of K .
We can describe all f ∈ Q[X ] for which f : Q → Q is (≥ 3)-to-1 over
infinitely many numbers. For instance, if f is (≥ 5)-to-1 infinitely often,
then f = g ◦ (X 3 − X )2 ◦ ` for some g , ` ∈ C[X ] with deg(`) = 1.
We cannot describe all f ∈ Q[X ] for which f : Q → Q is (≥ 2)-to-1 over
infinitely many points, and we do not believe that a reasonable description
is possible. The condition says that (f (X ) − f (Y ))/(X − Y ) = 0 has
infinitely many zeroes in Q × Q. When deg(f ) = 4, this asks which
members of a certain (infinite) collection of elliptic curves have positive
rank, a question much more difficult than the Birch–Swinnerton-Dyer
conjecture. (Bhargava will likely receive a Fields Medal for his proof that a
positive fraction of elliptic curves have rank zero; it is not believed that
there is a nice description of which curves these are.)
4 / 13
The exceptional polynomials
We can describe all f ∈ Q[X ] for which there is some number field K such
that f : K → K is (≥ 2)-to-1 over infinitely many elements of K .
We can describe all f ∈ Q[X ] for which f : Q → Q is (≥ 3)-to-1 over
infinitely many numbers. For instance, if f is (≥ 5)-to-1 infinitely often,
then f = g ◦ (X 3 − X )2 ◦ ` for some g , ` ∈ C[X ] with deg(`) = 1.
We cannot describe all f ∈ Q[X ] for which f : Q → Q is (≥ 2)-to-1 over
infinitely many points, and we do not believe that a reasonable description
is possible. The condition says that (f (X ) − f (Y ))/(X − Y ) = 0 has
infinitely many zeroes in Q × Q. When deg(f ) = 4, this asks which
members of a certain (infinite) collection of elliptic curves have positive
rank, a question much more difficult than the Birch–Swinnerton-Dyer
conjecture. (Bhargava will likely receive a Fields Medal for his proof that a
positive fraction of elliptic curves have rank zero; it is not believed that
there is a nice description of which curves these are.)
4 / 13
The exceptional polynomials
We can describe all f ∈ Q[X ] for which there is some number field K such
that f : K → K is (≥ 2)-to-1 over infinitely many elements of K .
We can describe all f ∈ Q[X ] for which f : Q → Q is (≥ 3)-to-1 over
infinitely many numbers. For instance, if f is (≥ 5)-to-1 infinitely often,
then f = g ◦ (X 3 − X )2 ◦ ` for some g , ` ∈ C[X ] with deg(`) = 1.
We cannot describe all f ∈ Q[X ] for which f : Q → Q is (≥ 2)-to-1 over
infinitely many points, and we do not believe that a reasonable description
is possible. The condition says that (f (X ) − f (Y ))/(X − Y ) = 0 has
infinitely many zeroes in Q × Q. When deg(f ) = 4, this asks which
members of a certain (infinite) collection of elliptic curves have positive
rank, a question much more difficult than the Birch–Swinnerton-Dyer
conjecture. (Bhargava will likely receive a Fields Medal for his proof that a
positive fraction of elliptic curves have rank zero; it is not believed that
there is a nice description of which curves these are.)
4 / 13
Mazur’s theorem and a conjectural generalization
Theorem (Mazur) Every elliptic curve over Q has at most 16 rational
torsion points.
Reformulation: If f : E1 → E2 is a nonconstant rational map between
genus-1 curves over Q, then the induced map E1 (Q) → E2 (Q) is at most
16-to-1.
Our result: If f : A1 → A1 is a finite morphism defined over Q, then the
induced map A1 (Q) → A1 (Q) is at most 6-to-1 outside a finite set.
Question: For any d, is there a constant Nd such that every morphism
f : V → W between d-dimensional varieties over Q induces a map
V (Q) → W (Q) which is at most Nd -to-1 outside a proper Zariski-closed
subset of W ?
This is not even known for d = 1, either for maps P1 → P1 or for maps
E → P1 . But it is known for maps A1 → A1 and E1 → E2 , and maps
P1 → P1 which come from maps E1 → E2 .
5 / 13
Mazur’s theorem and a conjectural generalization
Theorem (Mazur) Every elliptic curve over Q has at most 16 rational
torsion points.
Reformulation: If f : E1 → E2 is a nonconstant rational map between
genus-1 curves over Q, then the induced map E1 (Q) → E2 (Q) is at most
16-to-1.
Our result: If f : A1 → A1 is a finite morphism defined over Q, then the
induced map A1 (Q) → A1 (Q) is at most 6-to-1 outside a finite set.
Question: For any d, is there a constant Nd such that every morphism
f : V → W between d-dimensional varieties over Q induces a map
V (Q) → W (Q) which is at most Nd -to-1 outside a proper Zariski-closed
subset of W ?
This is not even known for d = 1, either for maps P1 → P1 or for maps
E → P1 . But it is known for maps A1 → A1 and E1 → E2 , and maps
P1 → P1 which come from maps E1 → E2 .
5 / 13
Mazur’s theorem and a conjectural generalization
Theorem (Mazur) Every elliptic curve over Q has at most 16 rational
torsion points.
Reformulation: If f : E1 → E2 is a nonconstant rational map between
genus-1 curves over Q, then the induced map E1 (Q) → E2 (Q) is at most
16-to-1.
Our result: If f : A1 → A1 is a finite morphism defined over Q, then the
induced map A1 (Q) → A1 (Q) is at most 6-to-1 outside a finite set.
Question: For any d, is there a constant Nd such that every morphism
f : V → W between d-dimensional varieties over Q induces a map
V (Q) → W (Q) which is at most Nd -to-1 outside a proper Zariski-closed
subset of W ?
This is not even known for d = 1, either for maps P1 → P1 or for maps
E → P1 . But it is known for maps A1 → A1 and E1 → E2 , and maps
P1 → P1 which come from maps E1 → E2 .
5 / 13
Mazur’s theorem and a conjectural generalization
Theorem (Mazur) Every elliptic curve over Q has at most 16 rational
torsion points.
Reformulation: If f : E1 → E2 is a nonconstant rational map between
genus-1 curves over Q, then the induced map E1 (Q) → E2 (Q) is at most
16-to-1.
Our result: If f : A1 → A1 is a finite morphism defined over Q, then the
induced map A1 (Q) → A1 (Q) is at most 6-to-1 outside a finite set.
Question: For any d, is there a constant Nd such that every morphism
f : V → W between d-dimensional varieties over Q induces a map
V (Q) → W (Q) which is at most Nd -to-1 outside a proper Zariski-closed
subset of W ?
This is not even known for d = 1, either for maps P1 → P1 or for maps
E → P1 . But it is known for maps A1 → A1 and E1 → E2 , and maps
P1 → P1 which come from maps E1 → E2 .
5 / 13
Mazur’s theorem and a conjectural generalization
Theorem (Mazur) Every elliptic curve over Q has at most 16 rational
torsion points.
Reformulation: If f : E1 → E2 is a nonconstant rational map between
genus-1 curves over Q, then the induced map E1 (Q) → E2 (Q) is at most
16-to-1.
Our result: If f : A1 → A1 is a finite morphism defined over Q, then the
induced map A1 (Q) → A1 (Q) is at most 6-to-1 outside a finite set.
Question: For any d, is there a constant Nd such that every morphism
f : V → W between d-dimensional varieties over Q induces a map
V (Q) → W (Q) which is at most Nd -to-1 outside a proper Zariski-closed
subset of W ?
This is not even known for d = 1, either for maps P1 → P1 or for maps
E → P1 . But it is known for maps A1 → A1 and E1 → E2 , and maps
P1 → P1 which come from maps E1 → E2 .
5 / 13
Proof strategy
Suppose f ∈ Q[X ] induces a map f : Q → Q which is (≥ 7)-to-1 over
infinitely many points.
Then the scheme V : f (X1 ) = f (X2 ) = · · · = f (X7 ) has an irreducible
component containing infinitely many rational points that do not lie in any
diagonal Xi = Xj (i 6= j).
Hence (Faltings) V has an irreducible component of genus 0 or 1.
Obstacle: it’s difficult to use this fact, since we don’t know how many
components V can have, or any formula for the genus of these
components.
We postpone addressing this issue by first classifying the f ’s which are
(≥ 2)-to-1 infinitely often.
6 / 13
Proof strategy
Suppose f ∈ Q[X ] induces a map f : Q → Q which is (≥ 7)-to-1 over
infinitely many points.
Then the scheme V : f (X1 ) = f (X2 ) = · · · = f (X7 ) has an irreducible
component containing infinitely many rational points that do not lie in any
diagonal Xi = Xj (i 6= j).
Hence (Faltings) V has an irreducible component of genus 0 or 1.
Obstacle: it’s difficult to use this fact, since we don’t know how many
components V can have, or any formula for the genus of these
components.
We postpone addressing this issue by first classifying the f ’s which are
(≥ 2)-to-1 infinitely often.
6 / 13
Proof strategy
Suppose f ∈ Q[X ] induces a map f : Q → Q which is (≥ 7)-to-1 over
infinitely many points.
Then the scheme V : f (X1 ) = f (X2 ) = · · · = f (X7 ) has an irreducible
component containing infinitely many rational points that do not lie in any
diagonal Xi = Xj (i 6= j).
Hence (Faltings) V has an irreducible component of genus 0 or 1.
Obstacle: it’s difficult to use this fact, since we don’t know how many
components V can have, or any formula for the genus of these
components.
We postpone addressing this issue by first classifying the f ’s which are
(≥ 2)-to-1 infinitely often.
6 / 13
Proof strategy
Suppose f ∈ Q[X ] induces a map f : Q → Q which is (≥ 7)-to-1 over
infinitely many points.
Then the scheme V : f (X1 ) = f (X2 ) = · · · = f (X7 ) has an irreducible
component containing infinitely many rational points that do not lie in any
diagonal Xi = Xj (i 6= j).
Hence (Faltings) V has an irreducible component of genus 0 or 1.
Obstacle: it’s difficult to use this fact, since we don’t know how many
components V can have, or any formula for the genus of these
components.
We postpone addressing this issue by first classifying the f ’s which are
(≥ 2)-to-1 infinitely often.
6 / 13
Proof strategy
Suppose f ∈ Q[X ] induces a map f : Q → Q which is (≥ 7)-to-1 over
infinitely many points.
Then the scheme V : f (X1 ) = f (X2 ) = · · · = f (X7 ) has an irreducible
component containing infinitely many rational points that do not lie in any
diagonal Xi = Xj (i 6= j).
Hence (Faltings) V has an irreducible component of genus 0 or 1.
Obstacle: it’s difficult to use this fact, since we don’t know how many
components V can have, or any formula for the genus of these
components.
We postpone addressing this issue by first classifying the f ’s which are
(≥ 2)-to-1 infinitely often.
6 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
We have an “if and only if” version of this result, which yields a
classification of uniqueness polynomials for meromorphic functions:
namely, f (X ) ∈ C[X ] such that, for nonconstant meromorphic functions A
and B, if f ◦ A = f ◦ B then A = B.
7 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
We have an “if and only if” version of this result, which yields a
classification of uniqueness polynomials for meromorphic functions:
namely, f (X ) ∈ C[X ] such that, for nonconstant meromorphic functions A
and B, if f ◦ A = f ◦ B then A = B.
7 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. If e.g. h2 = h3 (call it
h), then C(x1 , x2 , x3 ) is contained in the splitting field Ω of h(X ) − h(x1 )
over C(h(x1 )). So we compute the Galois group G of Ω/C(h(x1 )), and the
ramification in this extension, which lets us compute the genus of the
subfield of Ω fixed by the subgroup of G fixing any three prescribed roots
of h(X ) − h(x1 ). If the genus is ≤ 1, we consider C(x1 , x2 , x3 , x4 ), etc. 8 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. If e.g. h2 = h3 (call it
h), then C(x1 , x2 , x3 ) is contained in the splitting field Ω of h(X ) − h(x1 )
over C(h(x1 )). So we compute the Galois group G of Ω/C(h(x1 )), and the
ramification in this extension, which lets us compute the genus of the
subfield of Ω fixed by the subgroup of G fixing any three prescribed roots
of h(X ) − h(x1 ). If the genus is ≤ 1, we consider C(x1 , x2 , x3 , x4 ), etc. 8 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. If e.g. h2 = h3 (call it
h), then C(x1 , x2 , x3 ) is contained in the splitting field Ω of h(X ) − h(x1 )
over C(h(x1 )). So we compute the Galois group G of Ω/C(h(x1 )), and the
ramification in this extension, which lets us compute the genus of the
subfield of Ω fixed by the subgroup of G fixing any three prescribed roots
of h(X ) − h(x1 ). If the genus is ≤ 1, we consider C(x1 , x2 , x3 , x4 ), etc. 8 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. If e.g. h2 = h3 (call it
h), then C(x1 , x2 , x3 ) is contained in the splitting field Ω of h(X ) − h(x1 )
over C(h(x1 )). So we compute the Galois group G of Ω/C(h(x1 )), and the
ramification in this extension, which lets us compute the genus of the
subfield of Ω fixed by the subgroup of G fixing any three prescribed roots
of h(X ) − h(x1 ). If the genus is ≤ 1, we consider C(x1 , x2 , x3 , x4 ), etc. 8 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and h = h̃ ◦ ` with g , h̃, ` ∈ C[X ] such that ` is linear,
H(X , Y ) | [h(X ) − h(Y )], and h̃(X ) is either
a Chebyshev polynomial Tn (X )
X i (X + 1)j
r (X )n where deg(r ) ≤ 5
or a polynomial of degree at most 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. If e.g. h2 = h3 (call it
h), then C(x1 , x2 , x3 ) is contained in the splitting field Ω of h(X ) − h(x1 )
over C(h(x1 )). So we compute the Galois group G of Ω/C(h(x1 )), and the
ramification in this extension, which lets us compute the genus of the
subfield of Ω fixed by the subgroup of G fixing any three prescribed roots
of h(X ) − h(x1 ). If the genus is ≤ 1, we consider C(x1 , x2 , x3 , x4 ), etc. 8 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
From (≥ 2)-to-1 to (≥ 7)-to-1, continued
Theorem: For f (X ) ∈ C[X ] \ C, if H(X , Y ) is an irreducible factor of
f (X ) − f (Y ) such that the curve H(X , Y ) = 0 has genus ≤ 1, then
f = g ◦ h and there is a linear ` ∈ C[X ] such that h ◦ ` is either Tn (X ) or
X i (X + 1)j or r (X )n with deg(r ) ≤ 12.
To deduce our 6-to-1 result from this, let t be transcendental over C, and
let x1 , . . . , x7 be distinct roots of f (X ) − t such that C(x1 , . . . , x7 ) has
genus ≤ 1. Each pair (x1 , xk ) yields hk as above. On the previous slide we
determined the situations where h2 = h3 (for example). Now assume
h2 6= h3 . Note that f ∈ C(h2 ) ∩ C(h3 ). But Ritt showed that, if
u, v ∈ C[X ] satisfy C(u) ∩ C(v ) 6= C, then u = µ ◦ U ◦ R and
v = ν ◦ V ◦ R where µ, ν, R ∈ C[X ] with µ, ν linear and (U, V ) being
either (Tn , Tm ) or (X n , X i S(X n )) or (X i S(X n ), X n ). Most of our pairs
(h2 , h3 ) are not in any of Ritt’s forms.
These types of arguments restrict the candidates for an f which is
(≥ 7)-to-1 infinitely often. Compute ranks of elliptic curves, etc., to obtain
9 / 13
Infinitely non-injective polynomials: the irreducible case, I
We must classify genus-(≤ 1) factors of H(X , Y ) :=
f (X ) − f (Y )
.
X −Y
First suppose H is irreducible. Then the genus g of H = 0 can be expressed
in terms of the factorization types of all f (X ) − λ in C[X ] (with λ ∈ C):
If ef (α) denotes the multiplicity of α as a root of f (X ) − f (α), then
2g − 2 = −2(d − 1) +
X
(ef (α) − gcd(ef (α), ef (β))) .
α,β
f (α)=f (β)
Use this to determine all numerical plausibilities for the factorization types
of all f (X ) − λ, assuming H irreducible and g ∈ {0, 1}.
10 / 13
Infinitely non-injective polynomials: the irreducible case, I
We must classify genus-(≤ 1) factors of H(X , Y ) :=
f (X ) − f (Y )
.
X −Y
First suppose H is irreducible. Then the genus g of H = 0 can be expressed
in terms of the factorization types of all f (X ) − λ in C[X ] (with λ ∈ C):
If ef (α) denotes the multiplicity of α as a root of f (X ) − f (α), then
2g − 2 = −2(d − 1) +
X
(ef (α) − gcd(ef (α), ef (β))) .
α,β
f (α)=f (β)
Use this to determine all numerical plausibilities for the factorization types
of all f (X ) − λ, assuming H irreducible and g ∈ {0, 1}.
10 / 13
Infinitely non-injective polynomials: the irreducible case, I
We must classify genus-(≤ 1) factors of H(X , Y ) :=
f (X ) − f (Y )
.
X −Y
First suppose H is irreducible. Then the genus g of H = 0 can be expressed
in terms of the factorization types of all f (X ) − λ in C[X ] (with λ ∈ C):
If ef (α) denotes the multiplicity of α as a root of f (X ) − f (α), then
2g − 2 = −2(d − 1) +
X
(ef (α) − gcd(ef (α), ef (β))) .
α,β
f (α)=f (β)
Use this to determine all numerical plausibilities for the factorization types
of all f (X ) − λ, assuming H irreducible and g ∈ {0, 1}.
10 / 13
Infinitely non-injective polynomials: the irreducible case, I
We must classify genus-(≤ 1) factors of H(X , Y ) :=
f (X ) − f (Y )
.
X −Y
First suppose H is irreducible. Then the genus g of H = 0 can be expressed
in terms of the factorization types of all f (X ) − λ in C[X ] (with λ ∈ C):
If ef (α) denotes the multiplicity of α as a root of f (X ) − f (α), then
2g − 2 = −2(d − 1) +
X
(ef (α) − gcd(ef (α), ef (β))) .
α,β
f (α)=f (β)
Use this to determine all numerical plausibilities for the factorization types
of all f (X ) − λ, assuming H irreducible and g ∈ {0, 1}.
10 / 13
Infinitely non-injective polynomials: the irreducible case, II
Given a numerically plausible collection of factorization types of all
f (X ) − λ, the number of corresponding polynomials f (X ) can be counted
via computations in the fundamental group of a sphere minus several
points (thanks to Riemann’s existence theorem).
Given these counts, solve (infinitely many) differential equations to write
(Y )
down all such f , which finishes the case that f (XX)−f
is irreducible.
−Y
We cannot pass directly from the reducible case to the irreducible case,
(Y )
since factors of f (XX)−f
generally cannot be written in this form. Instead
−Y
we pass from the decomposable case to the indecomposable case, which
becomes the irreducible case via the following result:
Theorem (Fried, based on Schur) If f (X ) ∈ C[X ] is not a composition of
lower-degree polynomials, and is not a composition of linears with either
f (X ) − f (Y )
X n or Tn (X ), then
is irreducible.
X −Y
11 / 13
Infinitely non-injective polynomials: the irreducible case, II
Given a numerically plausible collection of factorization types of all
f (X ) − λ, the number of corresponding polynomials f (X ) can be counted
via computations in the fundamental group of a sphere minus several
points (thanks to Riemann’s existence theorem).
Given these counts, solve (infinitely many) differential equations to write
(Y )
down all such f , which finishes the case that f (XX)−f
is irreducible.
−Y
We cannot pass directly from the reducible case to the irreducible case,
(Y )
since factors of f (XX)−f
generally cannot be written in this form. Instead
−Y
we pass from the decomposable case to the indecomposable case, which
becomes the irreducible case via the following result:
Theorem (Fried, based on Schur) If f (X ) ∈ C[X ] is not a composition of
lower-degree polynomials, and is not a composition of linears with either
f (X ) − f (Y )
X n or Tn (X ), then
is irreducible.
X −Y
11 / 13
Infinitely non-injective polynomials: the irreducible case, II
Given a numerically plausible collection of factorization types of all
f (X ) − λ, the number of corresponding polynomials f (X ) can be counted
via computations in the fundamental group of a sphere minus several
points (thanks to Riemann’s existence theorem).
Given these counts, solve (infinitely many) differential equations to write
(Y )
down all such f , which finishes the case that f (XX)−f
is irreducible.
−Y
We cannot pass directly from the reducible case to the irreducible case,
(Y )
since factors of f (XX)−f
generally cannot be written in this form. Instead
−Y
we pass from the decomposable case to the indecomposable case, which
becomes the irreducible case via the following result:
Theorem (Fried, based on Schur) If f (X ) ∈ C[X ] is not a composition of
lower-degree polynomials, and is not a composition of linears with either
f (X ) − f (Y )
X n or Tn (X ), then
is irreducible.
X −Y
11 / 13
Infinitely non-injective polynomials: the irreducible case, II
Given a numerically plausible collection of factorization types of all
f (X ) − λ, the number of corresponding polynomials f (X ) can be counted
via computations in the fundamental group of a sphere minus several
points (thanks to Riemann’s existence theorem).
Given these counts, solve (infinitely many) differential equations to write
(Y )
down all such f , which finishes the case that f (XX)−f
is irreducible.
−Y
We cannot pass directly from the reducible case to the irreducible case,
(Y )
since factors of f (XX)−f
generally cannot be written in this form. Instead
−Y
we pass from the decomposable case to the indecomposable case, which
becomes the irreducible case via the following result:
Theorem (Fried, based on Schur) If f (X ) ∈ C[X ] is not a composition of
lower-degree polynomials, and is not a composition of linears with either
f (X ) − f (Y )
X n or Tn (X ), then
is irreducible.
X −Y
11 / 13
Infinitely non-injective polynomials: the irreducible case, II
Given a numerically plausible collection of factorization types of all
f (X ) − λ, the number of corresponding polynomials f (X ) can be counted
via computations in the fundamental group of a sphere minus several
points (thanks to Riemann’s existence theorem).
Given these counts, solve (infinitely many) differential equations to write
(Y )
down all such f , which finishes the case that f (XX)−f
is irreducible.
−Y
We cannot pass directly from the reducible case to the irreducible case,
(Y )
since factors of f (XX)−f
generally cannot be written in this form. Instead
−Y
we pass from the decomposable case to the indecomposable case, which
becomes the irreducible case via the following result:
Theorem (Fried, based on Schur) If f (X ) ∈ C[X ] is not a composition of
lower-degree polynomials, and is not a composition of linears with either
f (X ) − f (Y )
X n or Tn (X ), then
is irreducible.
X −Y
11 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Infinitely non-injective polynomials: the decomposable case
What if f = f1 ◦ f2 ◦ · · · ◦ fk ?
We get intermediate curves:
The genus can only stay the
same or increase as we progress
up through intermediate curves.
C
P1x
P1y
C0

P1f2 ◦···◦fk (x)
P1f2 ◦···◦fk (y )
f1
f1
P1f (x)

=
P1f (y )
Riemann-Hurwitz implies there
can be no ramification between
genus one curves.
Ramification at infinity usually
forces C to have genus greater
than one.
The case f1 = X n doubles the
length of the proof.
12 / 13
Another tool for the decomposable case
We often used a generalization of the following result:
Theorem
If f , g ∈ C[X ] cannot be written as compositions of lower-degree
polynomials, and f (X ) − g (Y ) is reducible in C[X , Y ], then f and g have
the same critical values: that is,
{f (a) : f 0 (a) = 0} = {g (b) : g 0 (b) = 0}.
This result can be generalized to any pair of maps between curves which
have the same target curve.
Conclusion: if f (X ) ∈ Q[X ] then only finitely many rational numbers have
more than six rational preimages under f .
13 / 13
Another tool for the decomposable case
We often used a generalization of the following result:
Theorem
If f , g ∈ C[X ] cannot be written as compositions of lower-degree
polynomials, and f (X ) − g (Y ) is reducible in C[X , Y ], then f and g have
the same critical values: that is,
{f (a) : f 0 (a) = 0} = {g (b) : g 0 (b) = 0}.
This result can be generalized to any pair of maps between curves which
have the same target curve.
Conclusion: if f (X ) ∈ Q[X ] then only finitely many rational numbers have
more than six rational preimages under f .
13 / 13
Another tool for the decomposable case
We often used a generalization of the following result:
Theorem
If f , g ∈ C[X ] cannot be written as compositions of lower-degree
polynomials, and f (X ) − g (Y ) is reducible in C[X , Y ], then f and g have
the same critical values: that is,
{f (a) : f 0 (a) = 0} = {g (b) : g 0 (b) = 0}.
This result can be generalized to any pair of maps between curves which
have the same target curve.
Conclusion: if f (X ) ∈ Q[X ] then only finitely many rational numbers have
more than six rational preimages under f .
13 / 13