Work with the exact pair factor
R = {(a,b) in Z_3 × Z_3 : a ≡ b mod 3}
≅ Z_3[eta]/(eta^2 - 3 eta).
Its maximal ideal is
m = 3(Z_3 × Z_3),
so
m^r = 3^r(Z_3 × Z_3)
for every r >= 1.
Because
m^r / m^(r+1) ≅ F_3^2
and the residue field R/m ≅ F_3 acts by scalars, every one-dimensional F_3-subspace of m^r/m^(r+1) lifts to an ideal.
So, for each finite quotient
R_k = R / 3^k R,
every ideal is one of two types:
a chain ideal m^r, 0 <= r <= k;
a line ideal
I_{r,lambda} = m^(r+1) + (Z/3^k Z)·v_{r,lambda},
for some 1 <= r < k and lambda in P^1(F_3).
A convenient choice of generators is
v_{r,0} = (3^r,0),
v_{r,1} = (3^r,3^r),
v_{r,2} = (3^r,-3^r),
v_{r,∞} = (0,3^r).
This gives exactly 4 intermediate ideals between m^r and m^(r+1) for every 1 <= r < k.
| quotient | number of chain ideals | number of line ideals | total ideals |
|---|---|---|---|
R/3R |
2 |
0 |
2 |
R/9R |
3 |
4 |
7 |
R/27R |
4 |
8 |
12 |
R/81R |
5 |
12 |
17 |
R/243R |
6 |
16 |
22 |
R/729R |
7 |
20 |
27 |
So the pair factor is not just a single radical chain. At every positive depth it opens into a full projective line P^1(F_3) of distinct line ideals.