For
R = Z_3[eta]/(eta^2 - 3 eta)
≅ {(a,b) in Z_3^2 : a ≡ b (mod 3)},
the CM generator is
M = B0 ⊕ R ⊕ B3,
with
B0 = R/(eta), B3 = R/(eta-3).
In the normalization model, write an element of R as a congruent pair (a,b). Then the fundamental maps are
q0(a,b) = a, q3(a,b) = b,
i0(x) = (3x,0), i3(x) = (0,3x),
eta(a,b) = (0,3b),
(3-eta)(a,b) = (3a,0).
So i0 and i3 are the two branch injections into the node, while q0,q3 are the two quotient maps out of the node.
These maps satisfy
q0 i0 = 3 id_B0,
q3 i3 = 3 id_B3,
i0 q0 = 3 - eta,
i3 q3 = eta,
q0 i3 = 0,
q3 i0 = 0,
eta^2 = 3 eta,
eta i0 = 0,
q0 eta = 0,
eta i3 = 3 i3,
q3 eta = 3 q3.
So the node loop eta kills the B0 branch, acts by multiplication by 3 on the B3 branch, and satisfies the same quadratic relation as the ring itself.
The irreducible-morphism skeleton is therefore
B0 <-> R <-> B3,
with generators
B0 --i0--> R --q0--> B0,
B3 --i3--> R --q3--> B3.
There are no direct cross-branch arrows, because
Hom(B0,B3) = Hom(B3,B0) = 0.
So the node is the unique mediator between the two branches, and its endomorphism loop eta records exactly how the two branch projections glue back into the singular order.
These relations were verified exhaustively modulo 81 in the explicit branch model.