SOLUTIONS TO THE FOURTH PROBLEM SHEET FOR
ALGEBRAIC NUMBER THEORY M3P15
AMBRUS PÁL
√
√
1. The norm of 1 + 2 −5 is 21 = 3 · 7, so (1 + 2 −5) is the product of prime ideals
of norm 3 and 7, respectively. Try
√
√
√
√
√
(3, 1 + 2 −5)(7, 1 + 2 −5) = (21, 3(1 + 2 −5), 7(1 + 2 −5), (1 + 2 −5)2 ) =
√
√
√
(21, 1 + 2 −5, (1 + 2 −5)2 ) = (1 + 2 −5).
√
We used that all terms are
√
√ divisible by 1 + 2 −5 in the second to last generator
and 1 + 2 −5, so
list. The norm of (3, 1 + 2 −5) must divide the norm of both 3 √
it divides 3. A similar argument shows that the norm of (7, √
1 + 2 −5) must divide
7. So the product above is the prime factorisation of (1 + 2 −5).
Note that
√
√
(3 + 2 −5, 3 − 2 −5) ⊇ (6),
√
so the norm of this ideal divides 36. It also divides
the √
norm of 3 + 2 −5, which is
√
the prime 29. Therefore the norm of (3 + 2 −5, 3 − 2 −5) divides 1, so it is the
unit ideal. Hence the√prime factorisation of this ideal is √
the empty product.
The norm of√7 + −5 is 54 = 2 · 33 . Note that 7 + −5 is not divisible by 3.
Therefore
7 + −5 must be the product of a prime ideal p1 of norm 2 containing
√
√
7 + −5, and the third power of a prime ideal p2 of norm 3 containing 7 + −5,
otherwise it would be divisible by the product
of a prime ideal of
√
√ norm 3 with its
conjugate, and hence by 3. Clearly (2, 7 + −5) ⊆ p1 and (3, 7 + −5) ⊆ p2 . Since
√
√
√
√
(2, 7 + −5)(2, 7 − −5) = (2, 1 + −5)(2, 1 − −5)
√
√
= (4, 2(1 + −5), 2(1 − −5), 6) = (2),
and
√
√
√
√
−5)(3, 7 − −5) = (3, 1 + −5)(3, 1 − −5) = (3),
√
√
we must have p1 = (2, 7 + −5) and p2 = (3, 7 + −5), and hence
√
√
√
(7 + −5) = (2, 7 + −5)(3, 7 + −5)3 .
√
2. (a) P = (2, 1 + −17).
(b) We need to show that P is not principal, since P 2 = (2) (easy to check by direct
computation), its order in the class group divides 2. Now P has norm 2, so if P
has one generator, then the norm of this generator is 2. Since x2 + 17y 2 = 2 has
no integral solutions, we conclude that P is not principal.
(c) Any Euclidean domain
not.
√ is a PID, so OK is√
(d) The norm of (1 + −17) is 18, so (1 + −17) = P QS where P is a√prime
ideal over 2, as above, and Q, S are prime ideals over 3. Trying with (3, 1 + −17)
(3, 7 +
Date: March 8, 2017.
1
2
Ambrus Pál
√
and (3, 1 − −17) we get ideals of norm
3. However QS is (3) so it will not work.
√
However using P Q2 we do get (1 + −17).
3. Since −23 ≡ 1 mod 4, by the Hasse–Minkowski bound every class in Cl(K) can
be represented by a non-zero ideal, whose norm is at most:
√
10
2 23
<
< 4.
λ(−23) =
π
3
Therefore every non-trivial ideal class can be represented by an ideal of norm 2 or
3. These must
be prime ideals.
In K the prime 2 is split, and the ideals of norm 2
√
√
1+ −23
1− −23
are (2,
) and (2,
), as
2
2
√
√
1 + −23
1 − −23
(2,
)(2,
)=
2
2
√
√
√
√
1 + −23
1 − −23
(4, 1 + −23, 1 − −23, 6) = (2, 2 ·
,2 ·
) = (2).
2
2
√
In K the prime 3 is also split, and the ideals of norm 3 are (3, 1+ 2−23 ) and
√
(3, 1− 2−23 ), as
√
√
√
√
1 + −23
1 − −23
1 + −23
1 − −23
(3,
)(3,
) = (9, 3 ·
,3 ·
, 6) = (3).
2
2
2
2
Since
√
√
√
1 + −23
1 + −23
1 + −23
(2,
)(3,
)=(
),
2
2
2
and by conjugation
√
√
√
1 − −23
1 − −23
1 − −23
(2,
)(3,
)=(
),
2
2
2
√
√
we get that Cl(K) is represented by the classes of (1), (2, 1+ 2−23 ) and (2, 1− 2−23 ),
so it has order at most 3.
4. The first statement is obvious. Every element of Cl(K) has a representative
which is an ideal I / OK . Write I = P1 · · · Pr , where the Pi are prime ideals. Since
P P = (p) or (p2 ), where P is a prime ideal over p, we have II = (a) for some a ∈ Z.
Hence the class of I in Cl(K) is the inverse of the class of I, so that the classes of
I and I in Cl(K) are equal if and only if the order of I in Cl(K) is at most 2.
5. By the above if the class of I has order at most 2 in Cl(K), then I = xI for
some x ∈ K. Write x = a/b, where a, b ∈ OK . Then aI = bI. Taking norms
and using the fact that Nr(I) = Nr(I) we obtain N (x) = ±1. The negative sign
is not possible because d < 0. By problem 8 in problem sheet 3 we can write
x = z/z, where z ∈ K. Let J = zI. This is a fractional ideal of K such that
J = J. Hence conjugate prime ideals in the decomposition of J have the same
power. As the product of conjugate ideals is generated by an element of Z, we get
that J = cP1 · · · Pr where Pi are distinct prime ideals over ramified primes, and
c ∈ Q∗ .
6. Let I = P1 · · · Pr , where Pi are distinct prime ideals over ramified primes with
r ≥ 1. If I = (a) where a ∈ OK , then N (a) = N (I) = p1 · · · pr , where Pi is the
unique prime ideal of OK over the prime pi . Note that p1 · · · pr divides d or 2d (the
last case occurs if d is 3 mod 4, and 2 is one of the p1 , . . . , pr ). When d = 1 the
ring OK is a PID, so all ideals are principal.
Solutions to the fourth problem sheet
3
If d < −1 is 2 or 3 mod 4, it is easy to check that there are no elements of norm
p1 · · · p√
r in OK unless p1 · · · pr = −d. In this last case the only element of norm
d is ± d. The case when
√ d is 1 mod 4 is similar. Thus the only possibility for I
to be principal is I = ( d), so I must be the product of all prime ideals over the
ramified primes when d is 2 mod 4, and of all ideals over odd ramified primes when
d is 1 or 3 mod 4.
7. The unique ideal P over a prime factor p|d is not principal, but P 2 = pOK is
principal.
8. Let d = −p1 · · · p2 , where p1 , . . . , pN are pairwise different prime √
numbers and
n > 2. To fix ideas assume that 2|d. Consider the√ideal Ji = (pi , d). This is
a prime ideal lying
over pi , which is ramified in Q( d). Hence √
Ji2 = p√i OK√, and
√
−1
−1
−1
−1
−1
Ji = pi √(pi , d). Thus we have Ji Jk = pk Ji Jk = pk (pi pk , p√i d, pk d, d)
√=
p−1
d),
because
(p
,
p
)
=
1.
I
claim
that
the
ideals
=
(p
,
d)
and
(p
p
,
d)
(p
p
,
i
k
i
i
j
i
k
k
for i 6= j are not principal.
Let
us
assume
this
for
a
moment.
Then
the
classes
[J
i]
√
in the class group of Q( d) are all non-trivial since Ji are not principal. These
classes are also pairwise distinct since Ji Jk−1 are not principal. This produces n
different non-trivial elements in √
the class group,√hence the result.
It remains to show that (pi , d) and (pi pj , d) are not principal. The norms
of these ideals are pi and pi pj , respectively. But there are no elements with such
norms in OK , since the equations x2 + |d|y 2 = pi and x2 + |d|y 2 = pi pj have no
integer solutions. Indeed, as |d| > pi pj , we get y = 0, and now it is clear that there
are no solutions (recall that pi is different from pj ). This completes the proof.
9. Since G is the product of its Sylow subgroups, and products of cyclic groups of
pair-wise relatively prime order is cyclic, we may assume that G is a p-group for
n−1
some prime p. Let the order of G be pn . Since the polynomial xp
− 1 = 0 as at
n−1
most p
roots, there is an x ∈ G whose order does not divide pn−1 . This x has
order pn , so it generates G.
10. First assume that R is a Dedekind domain and a UFD. Note that the product of
principal ideals is principal, and as R is Dedekind, every non-zero ideal is a product
of maximal ideals, so it will be enough to show the claim for maximal ideals. Now
let m /m R and pick a non-zero a ∈ m. Then we have a factorisation
(a) = p1 p2 · · · pn
with pi /m R. Then p1 p2 · · · pn ⊆ m so by one of the lemmas in the lectures one
of the ideals pi is m, say p1 = m. As R is also a UFD there is also a factorisation
a = uq1 q2 · · · qm with u ∈ R∗ and the qi are irreducible. Therefore
(a) = (q1 )(q2 ) · · · (qn ),
and as irreducible elements are prime in PID-s, the elements qi are prime, and
hence the ideals (qi ) are prime ideals. We get from the uniqueness of factorisation
into prime ideals that m = p1 = (qi ) for some index i, and so m is principal.
Now assume that R is a PID. Clearly every ideal in R is finitely generated, so
R is Noetherian. Since R is a UFD, it is also integrally closed. Now let I be a
non-zero prime ideal of R. Then I is generated by an element i which must be
prime. As i is irreducible, the ideal (i) is maximal among all principal ideals. But
all ideals of R are principal, so I = (i) is maximal.