Using the exact presentation
O_sing = Z_3 e7 ⊕ Z_3 epair ⊕ Z_3 n,
e7^2 = e7, epair^2 = epair, e7*epair = 0,
e7*n = 0, epair*n = n, n^2 = 3n,the Jacobson radical is
J = (3e7, 3epair, n).For every r >= 1, one gets the exact closed formula
J^r = Z_3·3^r e7 ⊕ Z_3·3^r epair ⊕ Z_3·3^(r-1) n.This immediately gives
O_sing / J ≅ F_3 e7 ⊕ F_3 epair (dimension 2),
J / 3 O_sing ≅ F_3·n (dimension 1),
O_sing / 3 O_sing has dimension 3.So the intrinsic finite residue split is exactly:
After that, the J-adic tower is infinite and every deeper graded piece has width 3.
| layer | generators of J^r |
dim J^r/J^(r+1) |
|---|---|---|
| 1 | 3^1 e7, 3^1 epair, n |
3 |
| 2 | 3^2 e7, 3^2 epair, 3^1 n |
3 |
| 3 | 3^3 e7, 3^3 epair, 3^2 n |
3 |
| 4 | 3^4 e7, 3^4 epair, 3^3 n |
3 |
| 5 | 3^5 e7, 3^5 epair, 3^4 n |
3 |
| 6 | 3^6 e7, 3^6 epair, 3^5 n |
3 |
If
alpha_r = [3^r e7],
beta_r = [3^r epair],
gamma_r = [3^(r-1) n],in J^r/J^(r+1), then
alpha_r alpha_s = alpha_(r+s),
beta_r beta_s = beta_(r+s),
alpha_r beta_s = 0,
alpha_r gamma_s = 0,
beta_r gamma_s = gamma_(r+s),
gamma_r gamma_s = gamma_(r+s).So the nilpotent residue class n mod 3 is only first-order nilpotent after reduction modulo 3; in the full J-adic algebra it propagates indefinitely through the relation n^2 = 3n.