CM / Torsion-Free Module Classification for the Pair Factor

Let

R = Z_3[eta]/(eta^2 - 3 eta)
  ≅ Z_3 ×_{F_3} Z_3.

Its normalization is

S = Z_3 × Z_3,

and the two branch modules are

B0 = R/(eta)      ≅ Z_3   (eta = 0),
B3 = R/(eta - 3)  ≅ Z_3   (eta = 3).

The node itself is the third basic module:

R.

Exact linear-algebra classification

Let M be a finitely generated torsion-free R-module. Suppose its ranks on the two branches are (a,b). After scaling inside the total quotient ring one may arrange

3(Z_3^a ⊕ Z_3^b) ⊆ M ⊆ Z_3^a ⊕ Z_3^b.

Set

W = M / 3(Z_3^a ⊕ Z_3^b) ⊂ F_3^a ⊕ F_3^b.

Define

K0 = W ∩ (F_3^a ⊕ 0),
K3 = W ∩ (0 ⊕ F_3^b).

Choose complements

F_3^a = K0 ⊕ U,
F_3^b = K3 ⊕ V.

Then the image of W in U ⊕ V has trivial intersections with the coordinate axes, so it is the graph of an isomorphism U -> V. After independent basis changes on U and V, that graph becomes a diagonal copy Δ^c of F_3^c.

Hence

W ≅ K0 ⊕ Δ^c ⊕ K3,

and therefore

M ≅ B0^u ⊕ R^c ⊕ B3^v

for unique integers

u = dim K0,   v = dim K3,   c = dim U = dim V.

So the indecomposable torsion-free / CM modules are exactly

B0,   R,   B3.

Explicit rank-2 verification

For Z_3-rank 2, the gluing space lives in F_3 ⊕ F_3. Up to the unit action (F_3^×)^2, the subspaces are:

orbit in F_3^2	representative(s)	resulting module	indecomposable?
zero	0	B0 ⊕ B3	no
left axis	<(1,0)>	B0 ⊕ B3	no
right axis	<(0,1)>	B0 ⊕ B3	no
diagonal	<(1,1)>, <(1,-1)>	R	yes
full plane	F_3^2	B0 ⊕ B3	no

Therefore there is no rank-2 indecomposable outside R.

Equivalently, the normalization module

S = Z_3 × Z_3

is not indecomposable; it splits as

S ≅ B0 ⊕ B3.

So the CM-category of the 3-adic Hecke node is finite and completely understood: three indecomposable objects, namely the two branches and the glued node.