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Algebra III
Home Assignment 3
Subhadip Chowdhury
Problem 9
Take any two elements x = a + bρ, y = c + dρ ∈ Z[ρ] where y 6= 0. Then we can write
x
a + bρ
(a + bρ)(c + dρ)
ac + bcρ + adρ + bd
ac + bd − ad
bc − ad
=
=
=
= 2
+ρ 2
2
2
2
y
c + dρ
(c + dρ)(c + dρ)
c + d − cd
c + d − cd
c + d2 − cd
where we have used the fact that ρ + ρ = −1 and ρρ = 1. Thus we can find rational numbers u, v such
that xy = u + vρ. Then we can find integers k, m such that |u − k| ≤ 12 and |v − m| ≤ 12 . Set p = u − k
and q = v − m. Thus x/y = (k + lρ) + (p + qρ). In other words,
x = (k + lρ)y + (p + qρ)y
Clearly, k + lρ ∈ Z[ρ]. So r = (p + qρ)y = x − (k + lρ)y ∈ Z[ρ]. Thus x = (k + lρ)y + r and for
N ≡ NmZ[ρ]/Z , we have,
N (r) = N ((p + qρ)y) = N (p + qρ)N (y) = (p2 − pq + q 2 )N (y)
Now
1
1
3
|p| ≤ , |q| ≤ ⇒ (p2 − pq + q 2 ) ≤ |p|2 + |p||q| + |q|2 ≤
2
2
4
Thus
3N (y)
< N (y)
4
Thus the function N defines a Euclidean algorithm on Z[ρ] making Z[ρ] an Euclidean ring.
0 ≤ N (r) ≤
If x = a + bρ ∈ Z[ρ] is a unit then N (x) = 1 implying that a2 − ab + b2 = 1. But a2 + b2 − ab ≥
ab ⇒ 1 ≥ ab. So either one of a, b is zero OR |a| = |b| = 1.
ˆ If a = 0, then b2 = 1 ⇒ b = ±1 ⇒ x = ±ρ.
ˆ If b = 0, then a2 = 1 ⇒ a = ±1 ⇒ x = ±1.
ˆ If a = b = 1 or a = b = −1 then x = ±(1 + ρ) = ∓ρ2 .
ˆ If a = −b = 1 or −a = b = 1 then a2 − ab + b2 6= 1; so this case is not possible.
Conversely note that
1.1 = −1. − 1 = ρ.(−1 − ρ) = −ρ.(1 + ρ) = ρ2 .ρ = −ρ2 .ρ = 1
Hence the units of Z[ρ] are ±1, ±ρ, ±ρ2 .
1
Algebra
Subhadip Chowdhury
Assignment 3
We claim that λ = 1 − ρ is a prime in Z[ρ]. If not, suppose λ = λ1 λ2 where λi ∈ Z[ρ] is not a unit.
Then N (λ1 )N (λ2 ) = N (λ) = N (1 − ρ) = 1 + 1 + 1 = 3. But N (λi ) ∈ Z ⇒ N (λi ) = 1 for some i. Wlog
suppose N (λ1 ) = 1. Then as we have seen in previous paragraph, λ1 is a unit. Contradiction! Thus λ is
a prime in Z[ρ]. Since Z[ρ] is an Euclidean domain, and hence a PID, (λ) is a maximal ideal.
Note that
λ2 = (1 − ρ)2 = 1 + ρ2 + ρ − 3ρ = −3ρ
Since −ρ is a unit, we thus have *******
First we prove the following claims
Claim 1: Every number in Z[ρ] is congruent to 0, 1, or − 1 mod λ.
Proof: Note that since λ = 1 − ρ, we have that ρ = 1 − λ. Hence given a + bρ ∈ Z[ρ], we have
a + bρ = a + b(1 − λ) = a + b − bλ
Hence a + bρ ≡ a + b mod λ. Since λ divides 3, our congruence is established.
Claim 2: Given α ∈ Z[ρ], λ divides the product α(α + 1)(α − 1).
Proof: It follows from Claim 1 that one of the three terms α, (α − 1) and (α + 1) is congruent to 0
mod λ.
Now take any θ 6∈ (λ) i.e λ does not divide θ. Then θ ≡ ±1 mod λ. Let θ = ±1 + αλ for some
α ∈ Z[ρ]. Hence modulo λ we have
θ3 ∓ 1 ≡(θ ∓ 1)(θ ∓ ρ)(θ ∓ ρ2 )
≡(αλ)(±1 + αλ ∓ ρ)(±1 + αλ ∓ ρ2 )
=(αλ)(±λ + αλ)(±1 + αλ ± 1 ± ρ)
=(αλ)(±λ + αλ)(±3 + αλ ∓ λ)
≡(αλ)(±λ + αλ)(±3 + αλ∓)λ(α ± 1)λ(α ∓ 1)λ
=λ3 (α)(α ± 1)(α ∓ 1)
But by Claim 2, we have (α)(α ± 1)(α ∓ 1) ≡ 0 mod λ. Hence
θ3 ∓ 1 ≡ 0
mod λ4 ⇐⇒ θ3 ≡ ±1
mod λ4
Suppose α, β, γ are coprime to λ. Then we have
α3 ≡ ±1
mod λ4 , β 3 ≡ ±1
mod λ4 , γ 3 ≡ ±1
mod λ4
Thus α3 + β 3 + γ 3 ≡ d mod λ4 where d ∈ Z, −3 ≤ d ≤ 3, d 6= 0. Note that if α3 + β 3 + γ 3 = 0 has
a solution then d ≡ 0 mod λ4 . But N (λ4 ) = N (λ)4 = 34 = 81 and N (d) ∈ {1, 4, 9}. Thus λ4 - d.
Contradiction!! Hence the equation α3 + β 3 + γ 3 = 0 does not have any solution.
2
Algebra
Subhadip Chowdhury
Assignment 3
Problem 10
Suppose L is a number field such that [L : Q] = n. Then we have seen in class that
LR ∼
= L ⊗Q R ∼
= OL ⊗Z R
Hence the quotient LR /OL is homeomorphic to a product of n circles, and in particular it is compact.
Note that we can canonically embed L ,→ LR . Thus we can consider OL as a discrete cocompact
subgroup of LR . Consider the open subsets Ut of LR defined by
Ut = {α ∈ LR : ∃β ∈ OL such that |tα − β| ≤ 1}
Now given any α ∈ LR , consider the sequence {nα}n∈N in LR . Let {nα}n∈N be the corresponding sequence
of the images under the natural projection map LR → LR /OL . Since LR /OL is compact, this sequence
has a convergent subsequence. In particular there is a Cauchy subsequence {nk α}k∈N . Thus given any
1 > > 0, for sufficiently large k and l,
|nk α − nl α| = |(nk − nl )α| < ⇒ ∃β ∈ OL such that |(nk − nl )α − β| < Thus α ∈ Unk −nl and thus
[
Ut = LR
t∈Z
In particular if U t denotes the image of Ut under the natural projection map LR → LR /OL , then we have
[
U t = LR /OL
t∈Z
Using compactness of LR /OL , we can find a finite subcover {U ti : 1 ≤ i ≤ n} which covers LR /OL . Thus
[
Uti = LR
1≤i≤n
So we can take
HL = max{N (ti ) : 1 ≤ i ≤ n}
and then all the conditions of the problem are satisfied.
Next we prove the following lemma:
Lemma: Given any ideal a ⊆ OL , ∃α ∈ a such that
|N (α)| ≤ HL |N (a)| = HL |OL /a|
3
(?)
Algebra
Subhadip Chowdhury
Assignment 3
Proof: Recall that any ideal in a Dedekind domain can be generated by at most two elements and by
our construction in problem 8, if α is any element in the ideal, we can find β such that α and β generate
the ideal. Now let α ∈ a − {0} be an element such that
|N (α)| = min |N (x)|
x∈a−{0}
Let a = ha, bi. Then since αβ ∈ L, we can find q ∈ OL such that t αβ − β ≤ 1. Thus by AM-Gm
inequality, we have
β
N t − β < 1 ⇒ |N (tβ − qα)| < |N (α)|
α
But that contradicts the minimality of |N (α)| since tβ − qα ∈ a unless tβ − qα = 0. Thus
q
β= α
t
Then
|N (α)|
#|OL /αOL |
=
=#
|OL /a|
#|OL /a|
a
αOL
=#
αOL + qt αOL
αOL
tαOL + qαOL
=#
tαO
L
OL
= |N (t)| = |t|n ≤ HL
≤#
tOL
Now consider the class of the ideal a, [a] as an element of cl(OL ). Then there exist a0 such that
[a0 ] = [a]−1 is inverse of [a] in cl(OL ). Also by above lemma there exists an element α ∈ a0 such that (?)
holds. Then consider the ideal b = αa0−1 ⊆ OL . Clearly [b] = [a]. Then by multiplicativity of Norm, we
get that
N (b) ≤ |N (α)|.|N (a)|−1 ≤ HL
√
√By a previous homework problem, we know that OL = Z[ −5].
√ Considering the embedding of
Q[ −5] in C, we can consider OL as the lattice generated by 1 and 5i.
√ √
Now given any α ∈ L, if α = a+ib, a, b ∈ Q, let α0 = α−bac−ibb/ 5c 5. Then
there exist t ∈ Z such
√ if √
that we can find β 0 ∈ OL with |α0 t−β 0 | < 1; then we can also find β = bact+ibb/ 5c 5t+β 0 ∈ OL so √
that
|tα − β| < 1. So it is enough to find a HL such that given any point α ∈ {a + ib : 0 ≤ a √
≤ 1, 0 ≤ b ≤
√ 5}
2
we can find a t with N (t) = |t| < HL such that |tα − β| < 1 for β = any one of 0, 1, 5i or 1 + 5i.
√
√
Draw discs of radius 1 around the four points 0, 1, 5i and 1 + 5i in C. Let us denote by S the
region of the rectangle R formed by those four points, which is covered by the four discs. Thus we need
to choose HL such that ∃t with N (t) < HL such that tα is in one of the four√discs. Looking at the
picture one can easily see that t = 1 or 2 suffices. Thus we can take HL = (1 + 5)2 .
Note that −5 6≡ 1 mod 4. Now
−5
2
1
=
=1
2
4
Algebra
Subhadip Chowdhury
−5
7
Assignment 3
1
−5
=
=1
3
3
−5
=0
5
2
=
= 1 since 7 ≡ 7
7
mod 8
Thus for p = 2, we can write
For p = 3,(7,
1
take c =
3
√
(2) = (p)2 where p = (2, −5 − 1)
√
we have (p) = pp where p = (p, −5 − c). Here c is a root of x2 + 5 mod p. Hence we can
if p = 3
Thus we have
if p = 7
(3) = (3, −1 +
√
√
−5)(3, −1 − −5)
and
√
√
(7) = (7, −3 + −5)(7, −3 − −5)
√
Also for p = 5, we have (5) = p2 where p = (p, −5 − c). Here c is a root of x2 + 5 mod 5 i.e. we may
write
√
√
(5) = (5, −5)2 = ( 5)2
Note that we have proved in second part every
√ 2ideal class in cl(OL ) has a representative ideal whose
ideal norm is less than
√ HL . Since HL = (1 + 5) , we need to only consider representative ideals with
norm less than (1 + 5)2 . Note the following two lemmas:
Lemma 1: Every prime ideal in OL divides a unique prime number. That is, if p is prime then p|(p)
for one prime p in N.
Proof: The ideal pp = (N (p)) is divisible by p and has a generator in N. Since p 6= (1), N (p) > 1.
Factor N (p) into primes in N say
N p = p1 p2 . . . pr
Then pp = (p1 p2 . . . pr ) = (p1 ) . . . (pr ), so p divides some (pi ). For the uniqueness, assume p|(p) and p|(q)
for two different prime numbers p and q. Then p ∈ p and q ∈ p. Since p and q are relatively prime, p
contains a pair of relatively prime integers, so p = (1). This is a contradiction.
Lemma 2: Every prime ideal in OL has norm p or p2 for some prime number p.
Proof: Let p be a prime ideal in OL . Then by lemma 1, there is a prime number p such that p|(p).
Taking ideal norms, N (p)|N ((p)). Since N ((p)) = |N (p)| = p2 , we have N (p) is p or p2 .
Thus using the value for HL found above and the
lemma we only need to consider prime ideals
√ above
2
such that their norms are prime numbers < (1 + 5) . Thus it is enough to consider prime ideals which
divide (2), (3), (5) and (7).
5
Algebra
Subhadip Chowdhury
Assignment 3
We want to find number of different ideal classes containing the ideals we calculated above; namely
√
√
√
√
( −5), (2, 1 − −5), (3, 1 ± −5), (7, 3 ± −5)
√
It is easy to see that (2, 1 + −5) is not principal. If it were, say (α) we would have
|N mL/Q (α)| = |N (α)| = 2
√
Hence N mL/Q (α) = ±2. Writing α = a + b −5, with a, b ∈ Z, we obtain a2 ± 5b2 = 2 which is
impossible.
√ Similarly we find that the prime divisors of (3) and (7) are not principal. The class containing
(2, 1 + −5) is an element of order 2 in cl(OL ). To investigate its relation with the classes containing
the primes
√ lying over 3, 7, we look for elements√whose norms are divisible by only 2, 3, 7. Considering
N (a √
+ b −5) = a2 + 5b2 , we notice that N (1 + −5) = 6, hence the prime factors
of the principal ideal
√
(1 + −5) lie over 2 and 3 since for each such prime factor p, N (p)|N√
((1 + −5)) = 6; hence N (p) = 2
or 3. Writing [I] for the ideal class containing I,√we find that [(2, 1 + −5)] is the inverse of√
[p] for one
of the prime ideals p lying over 3. Since [(2, 1 + −5)] has order 2, we find that [p] = [(2, 1 + −5)] and
it is easy to see that the same is true for the other prime q lying over 3(because √
q = p−1 ). A similar
argument shows that the primes lying over 7 are also on the same ideal class [(2, 1 + −5)]. We conclude
that all the nonprincipal ideals are in the same class. Thus there are exactly two ideal classes in cl(OL ).
Thus the ideal class number of cl(OL ) is 2.
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