Specializing the grading variable to one forgets graded decomposition information

The polynomials and the class $[P]$. In $\mathbb{Z}[q,q^{-1}]$ the elements $1+q^2$ and $2$ differ (their difference $q^2-1$ is nonzero) while $(1+q^2){\mathrm{\mid }}_{q=1}=2$; multiplying by the nonzero class $[D]$ gives distinct classes $[P]=(1+q^2)[D]$ and $2[D]$ in the free $A$-module $G_0^{\mathrm{g}\mathrm{r}}(H_{{\alpha }_0+{\alpha }_1}^{\mathrm{\Lambda }})$, whose images under the forgetful map $\zeta$ (which satisfies $\zeta (qm)=\zeta (m)$) both equal $2[D]$ Graded module and projective Grothendieck groups of the KLR and Hecke towers [F1].

The Specht classes and the transition polynomials. The example computes $[S(1,1)]=[D]$ and $[S(2)]=q[D]$, which are distinct because $q\mathrm{\ne }1$; the forgetful map sends both to $[D]$. The corresponding transition polynomials are $d_{(1,1),(1,1)}(q)=1$ and $d_{(1,1),(2)}(q)=q$ in the recorded $e=2$ Fock computation, distinct in $\mathbb{Z}[q,q^{-1}]$ and equal at $q=1$ (Projective and simple Grothendieck classes in a small non-semisimple block, Level-one global-basis transition polynomials in the symmetric-group convention) [F2].

Conclusion. By the graded decomposition theorem the classes are read through the transition polynomials, so the failure above is exactly the statement that the specialization $q\mapsto 1$ loses the graded multiplicities and shift data (Graded decomposition numbers equal canonical-basis coefficients in characteristic zero). The displayed finite block and polynomial computations introduce no new choice; AC is inherited only through the invoked Fock/global-basis and graded-decomposition identifications.