Let \(C_2=\langle g:g^2=1\rangle\). Over \({\mathbb Q}_2\), the two familiar sign projectors are
They do not belong to \({\mathbb Z}_2[C_2]\). Therefore one cannot first use them integrally and then reduce modulo two.
Compute the actual residual algebra
Over \({\mathbb F}_2\), put \(\delta=g-1=g+1\). Then \(\delta^2=g^2-2g+1=0\) and
The two sign eigenvalues have merged. The ring is local and retains a nilpotent direction instead of two residual projector pieces.
| Element | Square | Idempotent? |
|---|---|---|
| \(0\) | \(0\) | yes |
| \(1\) | \(1\) | yes |
| \(\delta\) | \(0\) | no |
| \(1+\delta\) | \(1\) | no |
In the basis \(1,\delta\), multiplication by \(g=1+\delta\) is
It is not the identity: \(G-I\) is nonzero and square-zero.
A trace can vanish while a packet is occupied
The identity projector on \({\mathbb F}_2^2\) has rank two and trace \(2=0\). Hence nonzero rank and nonzero trace are not equivalent in this characteristic. The localization and rank criteria still detect the nonzero space. The characteristic-zero qualification in Corollary 8.4 is essential.
Recovery without averaging
The augmentation map sends \(g\) to \(1\) and has kernel \((g-1)\). Thus
These quotient identities require no division by two. They explain why augmentation and coinvariants remain usable even when sign-projector formulas do not.
Read this as
Use the algebra available at the coefficient prime. Augmentation recovery is valid here; it still does not prove full deformation support or supply a universal automorphy theorem at two.