Solutions for Homework 1
1. First, we prove the rst formula by using proof by induction on n. When
n
n
n = 1, k=1 k 2 = n(n+1)(2n+1) = 1. Assume that k=1 k 2 = n(n+1)(2n+1)
6
6
is true. Then,
n+1
n
k 2 = (n + 1)2 +
k=1
k2
k=1
n(n + 1)(2n + 1)
6
(n + 1)(n

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2013
marques@cims.nyu.edu
Problem Set #5
Let prove rst a result extending the one of Z, if a =
p s are maximal ideal, then
min(ei ,fi )
a+b=
pi
i
max(ei ,fi )
and a b =
i
pei and b =
i
pi
i
pfi wher

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2013
marques@cims.nyu.edu
Problem Set #2
Due monday 16 September in Class
We recall the following important results good to know:
Let R be a GCD ring, and f (X) R[X]. Then the content of f , cont(f

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2013
marques@cims.nyu.edu
Problem Set #3
Exercise 2 p 23 [N]
Show that
3
54 = 2 3 =
13 +
47 13
47
2
2
are two dierent decomposition into irreducible integral elements of Q( 47).
Solution:
Let K =

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #4
Recall rst the two following result:
f
g
/L
/M
/N
/ 0 be a short exact sequence; then M is
Lemma: Let 0
Noetherian (Artinian) if and only if N and L are both

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #9
Exercise 2:
If L/K is a Galois extension of algebraic number elds, and P a prime ideal which is
unramied over K (i.e. p = P K is unramied in L), then there i

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #11
Exercise 1 p 71
Let A be an arbitrary ring, not necessarily an integral domain, let M be an A-module
and S a multiplicatively closed subset of A such that 0

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #8
Exercise 4 p 53
A prime ideal p of K is totally split in the separable extension L/K if and only if it is
totally split in the Galois closure N/K of L/K.
Sol

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #10
Recall that if f (x) Z[x] is a non-constant polynomial. Let Pf = cfw_p prime|n
N such that p|f (n) = 0. Then Pf is innite. Indeed, assume the contrary and

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #7
Exercise 1:
Let D > 1 be square free integer and d the discriminant of the real quadratic number
a
eld K = Q( D). Let x1 , y1 be the uniquely determined rati

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #1 with solution
Exercise 1 p 5 [N]:
Z[i] is unit if and only if N () = 1.
Solution:
Suppose that is a unit of Z[i] then there is Z[i] such the = 1, then N ()|

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #6
Recall that: a module P is projective if and only if for every surjective module homomorphism f : M P there exists a module homomorphism g : P M such that
f

Solutions for Homework 4
1. Lets consider the following situation: Alice wants to send a message to
Bob secretely. To do this, Bob needs to generate a public key and a private
key. First, he needs to pick two distinct prime numbers. Lets use p = 11
and q

Solutions for Homework 5
1. First, note that 231 = 3 7 11. By CRT, we need to consider:
5x3 93 0
mod 3
3
mod 7
3
mod 11
5x 93 2
5x 93 5
However,
5x3 0
3
5x 2
3
5x 5
mod 3 x 0
mod 3
3
mod 7 x 6
mod 7 x 3, 5, or 6
3
mod 11 x 1
mod 11 x 1
mod 7
mod 11
Thus,

Solutions for Homework 6
1. We have
z 4 = (x + iy )4 = x4 + 4ix3 y 6x2 y 2 4ixy 3 + y 4
1
z
z1
z+1
1
z2
= (x4 6x2 y 2 + y 4 ) + i(4x3 y 4xy 3 )
1
x iy
=
=2
x + iy
x + y2
y
x
i 2
=2
2
x +y
x + y2
x 1 + iy
(x 1 + iy )(x + 1 iy )
x2 (iy 1)2
=
=
=
x + 1 + iy

Solutions for Homework 7
1. Suppose that f is invertible i.e. there exists an arithmetic function g such
that f g = e. Then, f g (1) = f (1)g (1) = e(1) = 1, so we can conclude
that f (1) = 0. Conversely, assume that f (1) = 0. Dene an arithmetic
function

Solutions for Homework 8
1. (s) has simple poles at non positive integers s = 0, 1, 2, , and we
k
will prove that Res(s)s=k = (1) by proof by induction. Assume
k!
that k = 0. Note that (s + 1) = s(s) is true for any s, so we can
compute the residue in the

Solutions for Homework 9
1. Let g be a primitive element of (Z/pZ) . There are p 1 characters, and
for each k Z/(p 1)Z, the character k is given by
(n) =
0
e
2
if p | n
1lk
p1
if n g l
mod p
2. Let G be the set of all Dirichlet characters with the conduct

Dr. Marques Sophie
Oce 519
Number theory
Spring Semester 2014
marques@cims.nyu.edu
Problem Set #12
Exercise 1 p 84
Show that C[X, Y ]/(XY X), C[X, Y ]/(XY 1)., C[X, Y ]/(X 2 Y 3 ), C[X, Y ]/(Y 2
X 2 X 3 ) are one-dimensional noetherian rings. Which ones