Rank-one Hecke specializations can be semisimple or nonsemisimple

For any unit $q$, expanding [F1] and putting $S=qT$ gives $S^2=(q^2-1)S+q^2$, equivalently $(S-q^2)(S+1)=0$; thus the multiplicative source parameter is $\xi =q^2$, while all calculations below keep the local $T$ generator. Now set $q=i$. Then [F1] gives $(T-i)(T+i^{-1})=(T-i{)}^2=0$, and [F2] identifies this as the complete specialized presentation. With $ϵ=T-i$, translation $T=ϵ+i$ gives mutually inverse homomorphisms between $H_2(i)$ and $\mathbb{C}[ϵ]\mathrm{/}({ϵ}^2)$. Every class has a unique form $a+bϵ$: terms of degree at least two vanish, and a polynomial of degree at most one lying in $({ϵ}^2)$ has both coefficients zero. Hence the algebra has basis $1,ϵ$ and $ϵ$ is nonzero.

For $q=1$ and $q=-1$, [F1] gives the same relation $T^2-1=0$, using [F2] for the specialization. The elements $e_{+}=(1+T)\mathrm{/}2$ and $e_{-}=(1-T)\mathrm{/}2$ are orthogonal idempotents with sum $1$. The classes $1,T$ are linearly independent because no nonzero polynomial of degree at most one is divisible by $T^2-1$, so $e_{+}$ and $e_{-}$ are nonzero and form a basis. Thus $H_2(q)=\mathbb{C}{e}_{+}\oplus \mathbb{C}{e}_{-}$, and each one-dimensional left ideal is simple. The regular module is semisimple.

The left ideal $ϵH_2(i)$ is one-dimensional, nonzero and proper by step 1.1, so the left regular module is not simple. If it were a direct sum of simple left ideals, its dimension two would force it to be a sum of two one-dimensional simple modules. On any one-dimensional module, left multiplication by $ϵ$ is a scalar $\lambda$ satisfying ${\lambda }^2=0$, hence $\lambda =0$. It would then annihilate both summands and therefore all of $H_2(i)$, contradicting $ϵ\cdot 1=ϵ\mathrm{\ne }0$. Thus the regular module is not semisimple.

Steps 2.1 and 1.2 establish the two different specialization outcomes under the stated normalized relation. They make no claim about higher ranks, centralizers, positive characteristic, or every root of unity. No choice principle is used: every basis, idempotent, and module complement calculation is explicit and finite.