Math 627: Algebraic Number Theory
Final Assignment
April 29, 2011
Due Tuesday, May 10, 1:00pm
Instructions:
• Do all problems.
• You are permitted to take as much time as you need and to use the book, your class notes,
and homework. You should also feel to ask me questions.
• It is not permitted to use other sources, including other books or sources on the internet.
• If you would find it helpful, it is ok to use a software package (Maple, Mathematica, Sage,
etc.) to do routine calculations. More advanced procedures, such as built-in number theory
functions, are not to be used.
• If you have any questions about this assignment, please let me know.
Problems:
1. Let p be an odd prime, and let ζ = e2πi/p . For an integer a, let
Now let
p−1 X
a a
ζ ∈ Q(ζ).
η :=
p
a
p
be the Legendre symbol.
a=1
(a) For any embedding σ : Q(ζ) → C, show that σ(η) = ±η. Show that there exists an
embedding τ : Q(ζ) → C so that τ (η) = −η.
(b) Show that ηη = p, where η is the complex conjugate of η.
(c) Show that η 2 = (−1)(p−1)/2 p.
√
√
Remark 1: Thus η = ± p∗ , and so this gives a direct proof that Q( p∗ ) ⊆ Q(ζ).
√
√
2. Let K = Q( −3) with OK = Z[ω], where ω = e2πi/3 = − 12 + 2−3 .
(a) For m ≥ 1, let I(m) = #{a ⊆ OK : N (a) ≤ m}. Find a formula relating I(m) to the
number of points of OK inside a certain circle in the complex plain.
(b) Roughly speaking the number of lattice points in the circle can be approximated by the
area of the circle in (a) divided by the area of a fundamental domain for OK . Make this
precise and show that
π
I(m) = √ · m + e(m),
3 3
where e(m) is a function satisfying limm→∞
lim
m→∞
e(m)
m
= 0. And thus
I(m)
π
= √ .
m
3 3
(c) Show that for s > 1,
∞
X
ζK (s) =
m=1
1
1
−
,
I(m)
ms (m + 1)s
and then use (b) to deduce
π
lim (s − 1)ζK (s) = √ .
+
s→1
3 3
√
(d) Everything works out in pretty much the same way for K = Q( −D), D ∈ N squarefree,
√ except for one thing at the very beginning. What goes wrong in (a) for general
Q( −D)? What is the right formula for I(m) in terms of the number of points inside
the circle?
3. Let f (t) = t3 − 3t + 1 ∈ Z[t].
(a) Show that f is irreducible in Q[t].
(b) The intermediate value theorem reveals that f has 3 real roots, α < β < γ. Let
K = Q(γ). Show that both α, β ∈ K.
(c) Show that OK = Z[γ] and find the discriminant of K.
(d) For a prime number p ∈ Z, there are three possibilities for the factorization of the ideal
hpi ⊆ OK into the product of prime ideals. What are they?
(e) Express explicitly h2i, h3i, and h19i as the product of prime ideals in OK .
(f) Determine the class number of K.
√
√
√
4. Let K = Q( −5), and let a = h2, 1 + −5i, b = h3, 1 + −5i be ideals in OK .
(a) Show that neither a nor b is a principal ideal.
(b) Define a function
f : OK ⊕ O K → a ⊕ b
with the following properties:
• f is a group isomorphism under +.
• For all α, β, γ ∈ OK , f (γα, γβ) = γf (α, β).
(In other words, f is an isomorphism of OK -modules.)
Remark 1: Be careful. The natural map that first comes to mind for f is not surjective,
but it’s a good start.
Remark 2: In class we calculated that the class number of K is 2; what does this imply
about ab? In an oblique way, this property makes the construction of f possible.
Remark 3: Part (a) shows that a and b are not free OK -modules. Part (b) shows that
a ⊕ b is a free OK -module. Thus a and b are examples of finitely generated modules
which are projective but not free. This is not necessary to know to solve the problem,
but it is an interesting fact.