Embedding Quadratic Fields Into Quaternion Algebras

Home , Quadratic form

Embedding Quadratic Fields into Quaternion Algebras

Rose Mintzer-Sweeney, Alexander Schlesinger, Katherine Xiu

August 4, 2016

1 / 24 Overview

1 Background and Deﬁnitions

2 p-adic Analysis

3 Limits and a Conjecture

4 Sieving and Bounding

5 Further Questions

2 / 24 Quadratic Fields

Deﬁnition A quadratic ﬁeld is the set

√ √ Q( k) = {a + b k : a, b ∈ Q}

with the usual addition and multiplication. Here k is a squarefree integer other than 0 or 1.

3 / 24 i 2 = m, j2 = n, ij = −ji. Here m, n are again squarefree integers other than 0 or 1.

ij is often written as k.

Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by

4 / 24 j2 = n, ij = −ji. Here m, n are again squarefree integers other than 0 or 1.

ij is often written as k.

Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by i 2 = m,

4 / 24 ij = −ji. Here m, n are again squarefree integers other than 0 or 1.

ij is often written as k.

Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by i 2 = m, j2 = n,

4 / 24 Here m, n are again squarefree integers other than 0 or 1.

ij is often written as k.

Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by i 2 = m, j2 = n, ij = −ji.

4 / 24 ij is often written as k.

Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by i 2 = m, j2 = n, ij = −ji. Here m, n are again squarefree integers other than 0 or 1.

4 / 24 Rational Quaternion Algebras

Deﬁnition A quaternion algebra is the set of all elements

m, n = {a + bi + cj + dij : a, b, c, d ∈ Q} Q with component-wise addition and multiplication deﬁned by i 2 = m, j2 = n, ij = −ji. Here m, n are again squarefree integers other than 0 or 1.

ij is often written as k.

4 / 24 Example A more unusual example is 2,−3 . Here i 2 = 2, j2 = −3, and (ij)2 = 6. Q

Examples of Rational Quaternion Algebras

Example The classic example is −1,−1 , the Hamiltonian quaternions. Here Q i 2 = j2 = (ij)2 = −1.

5 / 24 Examples of Rational Quaternion Algebras

Example The classic example is −1,−1 , the Hamiltonian quaternions. Here Q i 2 = j2 = (ij)2 = −1.

Example A more unusual example is 2,−3 . Here i 2 = 2, j2 = −3, and (ij)2 = 6. Q

5 / 24 For ﬁxed k, we are interested in knowing for which (m, n) there is such an embedding.

Embeddings

Deﬁnition √ We say ( k) embeds into a quaternion algebra m,n if there exists an Q Q √ injective ring homomorphism φ : ( k) ,→ m,n . Q Q

6 / 24 Embeddings

Deﬁnition √ We say ( k) embeds into a quaternion algebra m,n if there exists an Q Q √ injective ring homomorphism φ : ( k) ,→ m,n . Q Q

For ﬁxed k, we are interested in knowing for which (m, n) there is such an embedding.

6 / 24 Example

√ −1, −1 Q( −1) → √ Q a + b −1 7→ a + bi

Example

√ 2, −3 Q( −1) → √ Q a + b −1 7→ a + b(i + j)

Examples of Embeddings

Let k = −1.

7 / 24 Example

√ 2, −3 Q( −1) → √ Q a + b −1 7→ a + b(i + j)

Examples of Embeddings

Let k = −1. Example

√ −1, −1 Q( −1) → √ Q a + b −1 7→ a + bi

7 / 24 Examples of Embeddings

Let k = −1. Example

√ −1, −1 Q( −1) → √ Q a + b −1 7→ a + bi

Example

√ 2, −3 Q( −1) → √ Q a + b −1 7→ a + b(i + j)

7 / 24 The embedding exists iﬀ ∃ω ∈ m,n such that ω2 = k. Q √ Given ω, the corresponding map is φ : a + b k → a + bω. The other direction is similar.

An Equivalent Condition

There is an equivalent, easier to check condition for the existence of φ.

8 / 24 √ Given ω, the corresponding map is φ : a + b k → a + bω. The other direction is similar.

An Equivalent Condition

There is an equivalent, easier to check condition for the existence of φ. The embedding exists iﬀ ∃ω ∈ m,n such that ω2 = k. Q

8 / 24 The other direction is similar.

An Equivalent Condition

There is an equivalent, easier to check condition for the existence of φ. The embedding exists iﬀ ∃ω ∈ m,n such that ω2 = k. Q √ Given ω, the corresponding map is φ : a + b k → a + bω.

8 / 24 An Equivalent Condition

There is an equivalent, easier to check condition for the existence of φ. The embedding exists iﬀ ∃ω ∈ m,n such that ω2 = k. Q √ Given ω, the corresponding map is φ : a + b k → a + bω. The other direction is similar.

8 / 24 Write ω = xi + yj + zij. We can check that

−k = −ω2 = mx2 + ny 2 + mnz2

Clearing denominators, we see that there exists ω such that ω2 = k iﬀ

kW 2 − mX 2 − nY 2 + mnZ 2 = 0

for some W , X , Y , Z ∈ Z not all zero.

Reduction to Quadratic Forms

One can check that for any ω such that ω2 = k, Re(ω) = 0.

9 / 24 −k = −ω2 = mx2 + ny 2 + mnz2

Clearing denominators, we see that there exists ω such that ω2 = k iﬀ

kW 2 − mX 2 − nY 2 + mnZ 2 = 0

for some W , X , Y , Z ∈ Z not all zero.

Reduction to Quadratic Forms

One can check that for any ω such that ω2 = k, Re(ω) = 0. Write ω = xi + yj + zij. We can check that

9 / 24 Clearing denominators, we see that there exists ω such that ω2 = k iﬀ

kW 2 − mX 2 − nY 2 + mnZ 2 = 0

for some W , X , Y , Z ∈ Z not all zero.

Reduction to Quadratic Forms

One can check that for any ω such that ω2 = k, Re(ω) = 0. Write ω = xi + yj + zij. We can check that

−k = −ω2 = mx2 + ny 2 + mnz2

9 / 24 kW 2 − mX 2 − nY 2 + mnZ 2 = 0

for some W , X , Y , Z ∈ Z not all zero.

Reduction to Quadratic Forms

One can check that for any ω such that ω2 = k, Re(ω) = 0. Write ω = xi + yj + zij. We can check that

−k = −ω2 = mx2 + ny 2 + mnz2

Clearing denominators, we see that there exists ω such that ω2 = k iﬀ

9 / 24 Reduction to Quadratic Forms

One can check that for any ω such that ω2 = k, Re(ω) = 0. Write ω = xi + yj + zij. We can check that

−k = −ω2 = mx2 + ny 2 + mnz2

Clearing denominators, we see that there exists ω such that ω2 = k iﬀ

kW 2 − mX 2 − nY 2 + mnZ 2 = 0

for some W , X , Y , Z ∈ Z not all zero.

9 / 24 Theorem (Hasse-Minkowski) Given an integer quadratic form Q(x), Q(x) = 0 has non-zero integer solutions if and only if Q(x) = 0 has non-zero real solutions, and

Q(x) = 0 has non-zero solutions in Zp for every prime p. Checking whether there are real solutions is easy.

Hasse-Minkowski

There are well-developed tools for handling a problem like this. We depend on two main theorems.

10 / 24 Q(x) = 0 has non-zero real solutions, and

Q(x) = 0 has non-zero solutions in Zp for every prime p. Checking whether there are real solutions is easy.

Hasse-Minkowski

There are well-developed tools for handling a problem like this. We depend on two main theorems. Theorem (Hasse-Minkowski) Given an integer quadratic form Q(x), Q(x) = 0 has non-zero integer solutions if and only if

10 / 24 Q(x) = 0 has non-zero solutions in Zp for every prime p. Checking whether there are real solutions is easy.

Hasse-Minkowski

10 / 24 Checking whether there are real solutions is easy.

Hasse-Minkowski

Q(x) = 0 has non-zero solutions in Zp for every prime p.

10 / 24 Hasse-Minkowski

Q(x) = 0 has non-zero solutions in Zp for every prime p. Checking whether there are real solutions is easy.

10 / 24 Theorem (Hensel’s Lemma) n Let P(x) be an integer polynomial. Suppose P(x0) = 0 mod p and 0 P (x0) 6= 0 mod p. Then ∃x˜0 ∈ Zp such that P(x ˜0) = 0.

So, we can lift solutions modulo p into solutions in Zp.

Hensel’s Lemma

Checking whether there are p-adic solutions is also tractable.

11 / 24 So, we can lift solutions modulo p into solutions in Zp.

Hensel’s Lemma

Checking whether there are p-adic solutions is also tractable. Theorem (Hensel’s Lemma) n Let P(x) be an integer polynomial. Suppose P(x0) = 0 mod p and 0 P (x0) 6= 0 mod p. Then ∃x˜0 ∈ Zp such that P(x ˜0) = 0.

11 / 24 Hensel’s Lemma

So, we can lift solutions modulo p into solutions in Zp.

11 / 24 Legendre Symbol

Recall that: Deﬁnition For an odd prime p and an integer x, the Legendre symbol is deﬁned by

 0 if p | x x  = 1 if ∃y s.t. x ≡ y 2 mod p . p −1 otherwise

12 / 24 1 (Conditions from R-analysis) If m < 0 and n < 0, then k < 0. 2 k (Conditions from Zp, p 6= 2) If p - k is an odd prime and p = 1, then −m n p · p If p | m, p | n, then p = 1.

n If p | m, p - n, then p = 1. m If p - m, p | n, then p = 1. 3 (Conditions from Z2) Omitted for the sake of brevity.

Conditions