Spencer Leonardis6-5-2015

Final HomeworkMath 111B

1. Prove if Q(x, y)= Ax2 +Bxy+Cy2 is a quadratic form, then the constant M = |A|+ |B|+ |C| satisfiesthe property that |Q(a,b)| É M ·max(|a|2, |b|2) for every a,b ∈R.Proof. If |a| > |b|, one has|Q(a,b)| = |Aa2 +Bab+Cb2| É |A||a|2 +|B||a|2 +|C||a|2 = (|A|+ |B|+ |C|)|a|2 = M|a|2 = M max(|a|2, |b|2),and if |a| < |b|, one has|Q(a,b)| = |Aa2 +Bab+Cb2| É |A||b|2 +|B||b|2 +|C||b|2 = (|A|+ |B|+ |C|)|a|2 = M|b|2 = M max(|a|2, |b|2). Ineither case, |Q(a,b)| É M ·max(|a|2, |b|2) for every a,b ∈R. 2

2. Prove that the class number of O−20 =Z[p−5] is 2.

Proof. We’ve shown in lecture that the ring Z[p−5] is not a unique factorization domain, and hence not

a principal ideal domain. In particular, the ideal J = (2,1+p−5) is not principal. Thus the class group isnot trivial and the class number C l

(Z[

p−5])

is > 1. If I (Z[p−5] is any non-principle ideal, let α ∈ I

have minimal nonzero norm, let β ∈ I \ ⟨α⟩ have minimal norm. Then β/α must lie in a region boundedby the unit circle, the lines x = 1

2 , x =−12 , and y= 1

2

p5. For any γ ∈ I with γ 6=β and γ/α in the same

region, β−γ is closer to one of 0,±α than the minimality of α allows. We conclude that I = ⟨α,β⟩.Observe that IJ is principal (i.e., by considering elements of small norm). Thus the class group hasprecisely two elements. 2

3. Prove the reciprocity law for the Kronecker symbol stated in class.Proof. Assume that M is odd and positive. Let m = pα1

1 · · · pαkk and M = qβ1

1 · · ·qβss be the prime

factorizations of m and M. Therefore, the Jacobi symbols(

Mm

)and

( mM

)are given by

(Mm

)=

k∏i=1

s∏j=1

( q j

pi

)and

( mM

)=

k∏i=1

s∏j=1

(pi

q j

). We have that

k∑i=1

= pi −12

≡ m−12

mod 2 ands∑

j=1= q j −1

2≡ M−1

2mod 2. Thus

k∑i=1

s∑j=1

pi −12

q j −12

≡ m−12

M−12

mod 2. By quadratic reciprocity,(

pi

q j

)( q j

pi

)= (−1)

pi−12

q j−12 . Thus

(Mm

)( mM

)=

k∏i=1

s∏j=1

(−1)pi−1

2q j−1

2 = (−1)

k∑i=1

s∑j=1

pi −12

q j −12 = (−1)

m−12

M−12 . 2

4. This is a new proof that a prime p ≡ 1 mod 4 can be written as a sum of two squares. Let u ∈Z satisfyu2 ≡−1 mod p (this exists since p ≡ 1 mod 4). Consider the lattice L ⊂R2 generated by (u,1) and (0, p).Show that any point (a,b) in L satisfies a2 +b2 ≡ 0 mod p. Calculate the area of a fundamentalparallelogram for L. Next, consider the open set E = {(x, y) ∈R2 : x2 + y2 < 2p}. Using the result proven inclass on convex symmetric open sets and lattices, prove that E contains a nonzero element (a,b) ∈ L.Conclude by showing that a2 +b2 = p.Proof. Let u ∈Z such that u2 ≡−1 mod p. We know that −1 is a quadratic residue modulo p if and onlyif p ≡ 1 mod 4. Let L be a lattice in R2, where L = {mv1+nv2 : m,n ∈Z} and v1 = (u,1) and v2 = (p,0). The

area A of a fundamental parallelogram of L is given by∣∣∣∣det

(1 k0 p

)∣∣∣∣= |p−0| = p. Let (x, y)= mv1 +nv2 for

some m,n ∈Z, i.e. (x, y) is a point on our lattice L. By definition of v1 and v2, it follows that x = mu+ pand y= m, and thus x2 + y2 = (mu+ p)2 +m2 = m2u2 +2mpu+ p2 +m2 = m2(u2 +1)+2mup+ p2. Since2mpu and p2 are both congruent to 0 modulo p, therefore m2(u2 +1)+2mpu+ p2 ≡ m2(u2 +1) mod p.Since u2+1≡ 0 mod p, and thus m2(u2+1)≡ 0 mod p. Let E be a circle centered symmetricically aboutthe origin with radius (2p)1/2. Then E = {(x, y) ∈R : x2 + y2 < 2p}. The area of E is given by 2πp, which is

1

greater than 4p = 4A. Therefore, E contains a lattice point other than the origin, arbitrarily given by(a,b). As (a,b) is a lattice point, a2 +b2 ≡ 0 mod p. Also, (a,b) ∈ E \ (0,0) and a2 +b2 < 2p. Since at leastone of a and b is nonzero, a2 +b2 Ê 0. Since 0< a2 +b2 < 2p, for a2 +b2 to be congruent to 0 modulo p, itmust be the case that a2 +b2 must be equal to p. Therefore every prime p ≡ 1 mod 4 is a sum of twosquares. 2

5. Let L(s)=∑∞n=1 an/ns be a Dirichlet series. Suppose that the partial sums of the coefficients

An =∑ni=1 ai are bounded, i.e. there exists a constant C such that |An| É C for all n. Then prove that L(s)

converges for s > 0, or more generally for all complex s ∈C with Re(s)> 0.

Proof. We assume that s = 0 (otherwise replace an by bn = an ·n−s). We prove the convergence of∞∑

n=1

anns

for Re(s)> 0 under the slightly weaker assumption that the sequence An =∑nk=1 ai of partial sums is

bounded. To prove the convergence of L(s) for Re(s)> 0, we check the Cauchy-criterion:∣∣∣∣∣ M∑n=K

an

ns

∣∣∣∣∣=∣∣∣∣∣ M∑n=K

An − An−1

nz

∣∣∣∣∣=

∣∣∣∣∣ AMMs − AK−1

K s +M−1∑n=K

An

(1ns −

1(n+1)s

)∣∣∣∣∣É |AM |Mx + |AK−1|

K x +M−1∑n=K

|An|∣∣∣∣ 1ns −

1(n+1)s

∣∣∣∣ ,

where x =Re(s). We have that1ns −

1(n+1)s = s

∫ n+1

n

dtts+1

and so ∣∣∣∣ 1ns −

1(n+1)s

∣∣∣∣É |s|∫ n+1

n

∣∣∣∣ 1ts+1

∣∣∣∣ dt = |s|∫ n+1

n

dttx+1 .

Noting that ∫ ∞

K

dttx+1 = 1

xK x ,

we see that ∣∣∣∣∣ M∑n=K

an

ns

∣∣∣∣∣É(2+ |s|

x

)1

K x ·supn

|An|.

Since K−x converges to 0 for K →∞ and supn

|Pn| <∞, it follows that

sK =K∑

n=1

an

nz

defines a Cauchy sequence, i.e.∞∑

n=1

an

nz

converges. (The convergence is locally uniform.) If∑

an converges absolutely, the absolute convergenceof

∑ annz for Re z Ê 0 follows from the comparison∣∣∣an

nz

∣∣∣= |an|nx É |an|,

and for bounded an, the absolute convergence of∑ an

nz for Re z > 1 follows by comparison with 1nx .

6. Let N be a positive integer, and let ψN be the trivial Dirichlet character with conductor N, soψN (a)= 0 if gcd(a, N) 6= 1 and ψN (a)= 1 if gcd(a, N)= 1. Prove that L(ψN , s)= ζ(s)

∏p|N (1− p−s).

Proof. We have the Euler product representation

L(ψN , s)= ∏p prime

(1− ψN (p)

ps

)−1

2

for Re s > 1, and the product converges absolutely, hence can be reordered and split arbitrarily. Splittingthe set of primes into those that divide N and those that don’t, we have

L(ψN , s)=(∏

p|N

(1− ψN (p)

ps

)−1)·( ∏

p-N

(1− ψN (p)

ps

)−1).

By definition of ψN , we have ψN (p)= 1 if p - N, and ψN (p)= 0 if p | N, so

L(ψN , s)=(∏

p|N

(1− 0

ps

)−1)·( ∏

p-N

(1− 1

ps

)−1)= ∏

p-N

(1− p−s)−1.

And we have

ζ(s)= ∏p prime

(1− p−s)−1

=( ∏

p-N

(1− p−s)−1

)·(∏

p|N

(1− p−s)−1

)= L(ψN , s) · ∏

p|N

(1− p−s)−1

= L(ψN , s)∏p|N

(1− p−s),

from which the desired

L(ψN , s)= ζ(s) · ∏p|N

(1− p−s)

follows by multiplication with the denominator of the right hand side. 2

3