A few class groups
The general strategy for determining the class group thus far has been as follows. Find explicit
representatives for ClK by using the theorem:
Theorem 0.1 Each x ∈ ClK can be represented by an ideal I ⊂ OK such that
p
N (I) ≤ (2/π)s |∆K |
where s is the number of conjugate pairs of complex imbeddings of K.
That is to say, the natural quotient map from the group of fractional ideals to ClK induces a surjection
p
{I : N (I) ≤ (2/π)s |∆K |}→ClK
Remark: This is not the best possible inequality, but good enough for our present purposes.
Here is the cubic example from class: K = Q(21/3 ).
We already know that Ok = Z[α] (where α = 21/3 ), ∆K = −27 × 4 = −108, and that
√
{I : N (I) ≤ (2/π) 108 ≈ 20/3 = 6. . . .}
surjects onto ClK under the projection map JK →ClK . The prime decomposition of such an I can’t
contain a prime ideal dividing any p > 5, so we need to find the primes lying above 2, 3 and 5 (we will
refer to prime ideals simply as primes, when the context makes the meaning clear). These are exactly
the primes that divide 2, 3, and 5. We have
x3 − 2 ≡ x3
mod 2, so that
(2) = P23
where P2 = (2, α) = (α). For 3, we have
x3 − 2 ≡ x3 + 1 ≡ (x + 1)3
mod 3, and hence,
(3) = P33
for P3 = (3, α + 1). Finally,
x3 − 2 ≡ (x − 3)(x2 + 3x − 1)
in F5 [x], so that
(5) = P5 P25
for P5 = (5, α − 3) and
P25 = (5, α2 + 3α − 1)
For the value of the norms,
N (P2 ) = 2, N (P3 ) = 3, N (P5 ) = 5, N (P25 ) = 25
(so we are indexing the primes by their norms). This follows from the fact that when P |p, P maximal,
then N (P ) = pf , where f is the degree of the irreducible factor of m̄(x) ∈ Fp [x] corresponding to P
(using the notation from class).
The possibilities for the I are then
OK , P2 , P3 , P5 , P22 , P2 P3
1
But P2 is principal, that is, gives the same class in ClK as OK , so we can delete it from the list and
just consider P3 and P5 . But
3 = α3 + 1 = (α + 1)(α2 − α + 1)
so that P3 = (α + 1). Thus, the only possible non-trivial element in ClK is represented by P5 . It turns
out that P5 is also principal. To show this, we need to find γ ∈ P5 such that N (γ) = 5. It is useful
here to recall the formula for norms: Whenever K = Q[β] and m(x) is the minimal polynomial of β,
then N (r −β) = m(r) for any r ∈ Q. To see this, note that in C[x], m(x) = (x−β1 )(x−β2 ) · · · (x−βd )
where the βi are the conjugates of β. But then,
m(r) = (r − β1 )(r − β2 ) · · · (r − βd ) = N (r − β)
For example,
N (α − 3) = −N (3 − α) = −(33 − 2) = −25
Since an ideal I is principal iff there is an a ∈ I such that |N (a)| = N (I), it is good to compute some
small norms of elements to compare with the norms of ideals. In this process, the preceding formuila
is quite useful. Carrying out this process just a litte bit will bring you to the norm of α + 2. This is
N (α + 2) = −N (−2 − a) = −m(−2) = 10
Since (α + 2) = (α − 3 + 5) ⊂ P5 , we have (α + 2) = P5 J for some ideal J. But then, our norm
computation says N (J) = 2. So the only choice for J is P2 , which we know is principal! Thus,
P5 = (α + 2)J −1 is principal. Since J = (α), we even know now that a generator for P5 is given by
(α + 2)/a = 1 + 2/α = 1 + α2 .
Consider now the case of K = Q[α] where α is a root of m(x) = x3 + 4x − 2. First, note that
0
m (x) = 3x2 + 4 > 0 for all x, hence, m(x) can have only one real root. So s = 1. The discriminant
for {1, α, α2 } is
∆K = −4 × 43 − 27 × 22 = −4 × 91
√
Hence, every class in ClK can be represented by an ideal I of norm ≤ (4/π) 91 ≈ 12.14. So we must
examine the primes lying above 2, 3, 5, 7, 11. We have
m(x) ≡ x3
so that
for P2 = (2, α).
mod 2
(2) = P23
m(x) ≡ (x − 1)(x2 + x + 1)
mod 3
so
(3) = P3 P9
with P3 = (3, α − 1) and P9 = (3, α2 + α − 1). m(x) mod 5 is irreducible, so that (5) = P5 is prime.
m(x) ≡ (x + 1)2 (x + 5)
so
mod 7
(7) = P72 Q7
where P7 = (7, α + 1) and Q7 = (7, α + 5).
m(x) ≡ (x − 5)(x2 + 5x + 7) (
so that
(11) = P11 P121
where P11 = (11, α − 5) and P121 = (11, α2 + 5α + 7).
2
mod 11)
Now, N (α) = −m(0) = 2 so that P2 = (α) is principal. N (1 − α) = m(1) = 3 so P3 is principal.
Since P3 P9 = (3) is principal, P9 is also principal. N (−1 − α) = m(−1) = −7 so P7 is principal.
Then Q7 is also principal since P72 Q7 = (7) is principal. Therefore, the only elements left to possibly
contribute non-trivially to ClK are P11 and P121 .
It is now useful to compute a few more simple norms:
N (2 − α) = 14, N (−2 − α) = −18, N (3 − a) = 37, N (−3 − α) = −41
N (4 − α) = 78, N (−4 − α) = −82, N (5 − α) = 143, N (−5 − α) = −147
At this point, we observe that 78 = 6 × 13 and 143 = 11 × 13.
This motivates us to consider the decomposition of (13), since this will give us a handle on the
decomposition of (4 − α) and (5 − α), thereby, more relations in the class group. We see already that
m(4) ≡ 0 mod 13 and m(5) ≡ 0 mod 13 and that
m(x) ≡ (x − 4)2 (x − 5)
So
mod 13
2
(13) = P13
Q13
where P13 = (13, α − 4) and Q13 = (13, α − 5). Meanwhile, because (α − 4) ⊂ P13 , by the previous
computation, the only choice for the prime decomposition of (α − 4) is (α − 4) = P13 P2 P3 , so that
−2
P13 must be principal. But then, Q13 = (13)P13
is also principal. Since (α − 5) = Q13 P11 is the only
possible decomposition of (α − 5), P11 is also principal.
Therefore, in conclusion
ClK = 1
√
√
K = Q[ −33]. Put α = −33. Since −33 ≡ 3 mod 4,we have that OK = Z[α]. √Also, ∆K =
−132 while s = 1. Thus, every ideal class has a representative I of norm N (I) ≤ (2/π) 132 ≈ 7.31.
So we need to examine the primes dividing 2, 3, 5, 7.
x2 + 33 ≡ (x + 1)2
so that
mod 2
(2) = P22
where P2 = (2, α + 1).
x2 + 33 ≡ x2
so
mod 3
(3) = P32
where P3 = (3, α).
x2 + 33 ≡ x2 − 2
mod 5
is irreducible, so (5) = P5 is prime.
x2 + 33 ≡ x2 − 2 ≡ (x − 3)(x + 3)
mod 7
so that
(7) = P7 Q7
where P7 = (7, α − 3) and Q7 = (7, α + 3). Since P5 is principal and hence redundant, we remove it
right away. When we write a general element in OK as m + nα, we get that N (m + nα) = m2 + 33n2 ,
from which we see right away that none of ±2, ±3, ±7 can be norms. Thus, P2 , P3 , P7 , Q7 are all nonprincipal. The other ideals of norm ≤ 7 are P22 , which is principal, and P2 P3 which is non-principal
(since 6 also can’t be a norm). Since P32 is principal, this shows that [P2 ] 6= [P3 ]. Clearly, [P2 P3 ] 6= [P2 ]
3
and [P2 P3 ] 6= [P2 ], since either equality will force one of P2 or P3 to be principal. Thus, we see that
{e, [P2 ], [P3 ], [P2 P3 ]} is an order 4 subgroup of of ClK in which each element has order 2. Hence, this
subgroup is isomorphic to Z/2 × Z/2. But we have aleady |ClK | ≤ 6, so that 4 ≤ |ClK | ≤ 6. The
possible finite abelian groups in this range of orders are
Z/4, Z/2 × Z/2, Z/5, Z/6
The only one with a subgroup isomorphic to Z/2 × Z/2 is Z/2 × Z/2 itself. That is,
ClK ' Z/2 × Z/2
(In particular, P7 and Q7 are each equivalent to one of the other elements.)
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