Download SOLUTIONS TO THE SECOND PROBLEM SHEET FOR

SOLUTIONS TO THE SECOND PROBLEM SHEET FOR
ALGEBRAIC NUMBER THEORY M3P15
AMBRUS PÁL
1. Note that
ω2 =
√
√
1
1
(1 − 3 − i2 3) = (−1 − i 3) = ω = −1 − ω,
4
2
and hence
ω 3 = ωω = 1.
Clearly E is closed under addition, and it contains 0 and 1, so we only need to show
that it is closed under multiplication. We compute:
(a + bω)(x + yω) = ax + (ay + bx)ω + byω 2 = (ax − by) + (ay + bx − by)ω.
2. Note that:
(a + bω)(a + bω) = (a + bω)(a + bω) = a2 + b2 + ab(ω + ω) = a2 + b2 − ab.
In particular for every a + bω ∈ E the number N (a + bω) is positive, and therefore
the function N is well-defined. Since it is the restriction of the complex norm onto
E, it is multiplicative, too.
3. For one of the three triangles ABP , BCP or ACP the angle at P is at least
120 degrees. We may assume that this holds for the triangle ABP without the
loss of generality. Let a, b, and c denote the length of the sides AP , BP or AB,
respectively. Let α denote the angle at P . By the cosine law:
1 = c2 = a2 + b2 − 2ab cos(α) ≥ a2 + b2 + ab
since α is at least 120 degrees. We may assume that a ≥ b without the loss of
generality. Then
1 ≥ 3b2 .
4. It will be enough to show that for every x, y ∈ E − {0} such that x does not
divide y there is a q ∈ E such that N (r) < N (x) where r = y − xq. Let z = y/x.
Since on the complex plane the Eisenstein integers are the vertices of a mosaic
which consists
of triangles whose sides have length one, there is a q ∈ E such that
√
3
|z − q| ≤ 3 by the previous problem. For this choice of q we get
N (r) = |r|2 = |y − xq|2 = |z − q|2 |x|2 ≤
1
N (x).
3
5. This is Theorem 1.14 in the notes.
6. Note that u = a + bω ∈ E is a unit if and only if N (a + bω) = 1. Since
b
3
1 = N (u) = a2 + b2 − ab = (a − )2 + b2 ,
2
4
Date: February 13, 2017.
1
2
AMBRUS PÁL
we get |b| ≤ 1. If b = 0 then a = ±1. If b = ±1 then a − 2b = ± 12 . Hence either
a = 0 or a = b = 1 or a = b = −1. These numbers are all units.
7. If x, y ∈ Z are solutions to the Diophantine equation:
x2 + 3y 2 = n
then set a = x + y and b = 2y. We compute:
a2 + b2 − ab = (x + y)2 + 4y 2 − 2(x + y)y = x2 + 3y 2 = n,
therefore if the second equation has a solution, so does the first. Now let a, b ∈ Z
be solutions to the Diophantine equation:
a2 + b2 − ab = n.
Then we have:
3
a
3
a+b 2
a−b 2
b
) + 3(
) .
n = (a − )2 + b2 = (b − )2 + a2 = (
2
4
2
4
2
2
Therefore if either a or b is even, the either (b − a/2, a/2) or (a − b/2, b/2) is an
integer solution for the second equation in the problem. If both of them are odd,
then the pair ((a + b)/2, (a − b)/2) is a solution for the second equation.
8. Note that as a group F∗p is isomorphic to Z/(p − 1)Z, so the first two conditions
are equivalent. If ω is a non-trivial 3-rd root of unity in Fp , then = 1 + 2ω is a
square-root of −3. Similarly, if is a square-root of −3 in Fp , then ω = (−1 + )/2
is a nontrivial 3-rd root of unity. Therefore the last two conditions are equivalent,
too.
9. Because E is a unique factorisation domain, a prime number p ∈ N remains a
prime in E if and only if it is not a norm of an element of E. By problem 6 this
happens if and only if the Diophantine equation:
x2 + 3y 2 = p
has no solution. This is not the case when p = 3, so we may assume that p 6= 3.
The congruence
x2 + 3y 2 ≡ x2 ≡ p mod 3
has a solution if and only if p ≡ 1 mod 3. Therefore p is a prime if p ≡ −1 mod 3.
On the other hand if p ≡ 1 mod 3 is a prime then by problem 7 the congruence
x2 ≡ −3 mod p has a solution.
Let √
x be such a solution; then p divides one of the
√
Eisenstein integers x + i 3 or x − i 3, since it divides their product. This implies
that p divides ±2 which is a contradiction.
10. This Diophantine equation has a solution if and only if for every prime number
p ∈ N such that p ≡ −1 mod 3 has an even exponent in the prime factorisation of
n.