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Embedding Quadratic Fields into Quaternion Algebras

Rose Mintzer-Sweeney, Alexander Schlesinger, Katherine Xiu

August 4, 2016

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Overview

1 Background and Definitions

2 p-adic Analysis

3 Limits and a Conjecture

4 Sieving and Bounding

5 Further Questions

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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 integerother than 0 or 1.

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Rational Quaternion Algebras

Definition

A quaternion algebra is the set of all elements

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

with component-wise addition and multiplication defined by

i2 = m,

j2 = n,

ij = −ji .

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

ij is often written as k .

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Definition

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

i2 = m,

j2 = n,

ij = −ji .

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Definition

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

i2 = m,

j2 = n,

ij = −ji .

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Definition

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

i2 = m,

j2 = n,

ij = −ji .

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Definition

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

i2 = m,

j2 = n,

ij = −ji .

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Definition

)= {a + bi + cj + dij : a, b, c , d ∈ Q}

i2 = m,

j2 = n,

ij = −ji .

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Examples of Rational Quaternion Algebras

Example

The classic example is(−1,−1

), the Hamiltonian quaternions. Here

i2 = j2 = (ij)2 = −1.

Example

A more unusual example is(2,−3Q

). Here i2 = 2, j2 = −3, and (ij)2 = 6.

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Examples of Rational Quaternion Algebras

Example

The classic example is(−1,−1

), the Hamiltonian quaternions. Here

i2 = j2 = (ij)2 = −1.

Example

A more unusual example is(2,−3Q

). Here i2 = 2, j2 = −3, and (ij)2 = 6.

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Embeddings

Definition

We say Q(√k) embeds into a quaternion algebra

)if there exists an

injective ring homomorphism φ : Q(√k) ↪→

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

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Embeddings

Definition

We say Q(√k) embeds into a quaternion algebra

)if there exists an

injective ring homomorphism φ : Q(√k) ↪→

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

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Examples of Embeddings

Let k = −1.

Example

Q(√−1)→

(−1,−1

)a + b

√−1 7→ a + bi

Example

Q(√−1)→

(2,−3

)a + b

√−1 7→ a + b(i + j)

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Let k = −1.

Example

Q(√−1)→

(−1,−1

)a + b

√−1 7→ a + bi

Example

Q(√−1)→

(2,−3

)a + b

√−1 7→ a + b(i + j)

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Let k = −1.

Example

Q(√−1)→

(−1,−1

)a + b

√−1 7→ a + bi

Example

Q(√−1)→

(2,−3

)a + b

√−1 7→ a + b(i + j)

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An Equivalent Condition

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

The embedding exists iff ∃ω ∈(m,nQ

)such that ω2 = k .

Given ω, the corresponding map is φ : a + b√k → a + bω.

The other direction is similar.

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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 + ny2 + 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.

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−k = −ω2

= mx2 + ny2 + mnz2

kW 2 −mX 2 − nY 2 + mnZ 2 = 0

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−k = −ω2

= mx2 + ny2 + mnz2

kW 2 −mX 2 − nY 2 + mnZ 2 = 0

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−k = −ω2

= mx2 + ny2 + mnz2

kW 2 −mX 2 − nY 2 + mnZ 2 = 0

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−k = −ω2

= mx2 + ny2 + mnz2

kW 2 −mX 2 − nY 2 + mnZ 2 = 0

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Hasse-Minkowski

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

Theorem (Hasse-Minkowski)

Given an integer quadratic form Q(x), Q(x) = 0 has non-zero integersolutions 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.

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Hasse-Minkowski

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Hasse-Minkowski

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Hasse-Minkowski

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Hasse-Minkowski

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Hensel’s Lemma

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

Theorem (Hensel’s Lemma)

Let P(x) be an integer polynomial. Suppose P(x0) = 0 mod pn andP ′(x0) 6= 0 mod p. Then ∃x0 ∈ Zp such that P(x0) = 0.

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

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Hensel’s Lemma

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Hensel’s Lemma

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Legendre Symbol

Recall that:

Definition

For an odd prime p and an integer x , the Legendre symbol is defined by

0 if p | x1 if ∃y s.t. x ≡ y2 mod p

−1 otherwise

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Conditions

Theorem (Ehrman, Mintzer-Sweeney, Schlesinger, Sheydvasser, Xiu2016)

There exists an embedding Q(√k) ↪→

)iff the following conditions

are met

1 (Conditions from R-analysis) If m < 0 and n < 0, then k < 0.

2 (Conditions from Zp, p 6= 2) If p - k is an odd prime and(kp

If p | m, p | n, then

(−mp · npp

If p | m, p - n, then(

If p - m, p | n, then(

3 (Conditions from Z2) Omitted for the sake of brevity.

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Conditions

are met

(−mp · npp

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Conditions

are met

(−mp · npp

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Conditions

are met

(−mp · npp

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Conditions

are met

(−mp · npp

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Conditions

are met

(−mp · npp

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Conditions

are met

(−mp · npp

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Long-Term Behavior

The next logical question is to determine how often these conditions aremet. Define

(m, n) ∈ Z2 : 1 < m, n < N, m, n squarefree}

Conjecture

For any fixed k ,

limN→∞

#(SN ∩

{(m, n) : Q

(√k)↪→(m,nQ

)})#SN

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Long-Term Behavior

The next logical question is to determine how often these conditions aremet. Define

(m, n) ∈ Z2 : 1 < m, n < N, m, n squarefree}

Conjecture

For any fixed k ,

limN→∞

#(SN ∩

{(m, n) : Q

(√k)↪→(m,nQ

)})#SN

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Our Theorem

Theorem (EMSSSX 2016)

For any fixed k,

limN→∞

#(S ′N ∩

{(m, n) : Q

(√k)↪→(m,nQ

)})#S ′N

S ′N ={

(m, n) ∈ Z2 : 1 < m, n < N, m, n squarefree and coprime}.

Note: we are using 1 < m, n < N for simplicity, but the same proof appliesto −1 > m, n > −N.

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Our Theorem

Theorem (EMSSSX 2016)

For any fixed k,

limN→∞

#(S ′N ∩

{(m, n) : Q

(√k)↪→(m,nQ

)})#S ′N

S ′N ={

(m, n) ∈ Z2 : 1 < m, n < N, m, n squarefree and coprime}.

Note: we are using 1 < m, n < N for simplicity, but the same proof appliesto −1 > m, n > −N.

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Our Approach

How can one prove a result like this?

One can check that #S ′N ≈(

)3N2 for large N.

To show that the limit is 0, we find an upper bound for the numeratorthat is o(N2).

We do this by focusing exclusively on the p-adic conditions on ncoming from m.

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Our Approach

)3N2 for large N.

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Our Approach

)3N2 for large N.

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Our Approach

)3N2 for large N.

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Definitions for a Sieve Approach

Recall that there is an embedding Q(√k) ↪→

)only if

If p | m, p - n, and

)= 1, then

Letm =

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Recall that there is an embedding Q(√k) ↪→

)only if

If p | m, p - n, and

)= 1, then

Letm =

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Definition

The admissible class is the set

A(m) := {x ∈ Z/mZ :

)= 1, ∀p | m}

Note that n satisfies the conditions for an embedding if and only if

n mod m ∈ A(m)

Thus an upper bound for the numerator of our limit is given by∑(m,n)∈S ′

1{n mod m∈A(m)}

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Definition

A(m) := {x ∈ Z/mZ :

)= 1, ∀p | m}

n mod m ∈ A(m)

1{n mod m∈A(m)}

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Definition

A(m) := {x ∈ Z/mZ :

)= 1, ∀p | m}

n mod m ∈ A(m)

1{n mod m∈A(m)}

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Approximating the Size of the Admissible Class

This upper bound is useful because we have a good approximation of thesize of the admissible class A(m).

Specifically,

#A(m) ≈ m

2ω(m),

where ω(x) is the number of distinct prime factors of x .

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Approximating the Size of the Admissible Class

This upper bound is useful because we have a good approximation of thesize of the admissible class A(m). Specifically,

#A(m) ≈ m

2ω(m),

where ω(x) is the number of distinct prime factors of x .

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An Upper Bound

∑(m,n)∈S ′

1{n mod m∈A(m)} ≈∑m<N

m(#A(m))

≈ N∑m<N

2ω(m)

Therefore, if we show that

∑m<N

2ω(m)= o(N),

then we are done.

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An Upper Bound

∑(m,n)∈S ′

1{n mod m∈A(m)} ≈∑m<N

m(#A(m))

≈ N∑m<N

2ω(m)

Therefore, if we show that

∑m<N

2ω(m)= o(N),

then we are done.

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Splitting the Sum

Note that, on average, ω(x) ≈ log log(x).

So, we split our sum into two pieces: one where ω is close to theaverage, and one where it is not.

Definition

B(N) =

{m < N :

∣∣∣∣ω(m)− 1

2loglog(N)

∣∣∣∣ < loglog(N)

}B(N)c = {m < N : m /∈ B(N)}

∑m<N

2ω(m)=

∑m∈B(N)

2ω(m)+

∑m∈B(N)c

2ω(m).

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Splitting the Sum

Definition

B(N) =

{m < N :

∣∣∣∣ω(m)− 1

2loglog(N)

}B(N)c = {m < N : m /∈ B(N)}

∑m<N

2ω(m)=

∑m∈B(N)

2ω(m)+

∑m∈B(N)c

2ω(m).

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Splitting the Sum

Definition

B(N) =

{m < N :

∣∣∣∣ω(m)− 1

2loglog(N)

}B(N)c = {m < N : m /∈ B(N)}

∑m<N

2ω(m)=

∑m∈B(N)

2ω(m)+

∑m∈B(N)c

2ω(m).

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Splitting the Sum

Definition

B(N) =

{m < N :

∣∣∣∣ω(m)− 1

2loglog(N)

}B(N)c = {m < N : m /∈ B(N)}

∑m<N

2ω(m)=

∑m∈B(N)

2ω(m)+

∑m∈B(N)c

2ω(m).

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An Outline of How to Control the Sums

Why is this splitting useful?

B(N) is tightly controlled, so it is easy to show that the sum overB(N) is o(N).

We can use a result of Granville and Soundararajan (2006)1 to showthat #B(N)c is small.

It follows that the sum over B(N)c is also o(N).

We are done!

1A. Granville and K. Soundararajan, Sieving and the Erdos-Kac Theorem22 / 24

We are done!

Further Research

How fast does this limit goes to 0?

How small of an upper bound on the number of embeddings can weobtain using sieve methods?Can we find a lower bound on the number of embeddings?

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Further Research

How small of an upper bound on the number of embeddings can weobtain using sieve methods?

Can we find a lower bound on the number of embeddings?

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Further Research

How small of an upper bound on the number of embeddings can weobtain using sieve methods?Can we find a lower bound on the number of embeddings?

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Acknowledgements

We would like to thank Max Ehrman and Senia Sheydvasser for theirwonderful mentorship and expert guidance.We would like to thank Sam Payne, Jose Gonzalez, and Michael Magee fororganizing the SUMRY program.Finally, we would like to thank MathFest for having us!

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