A perfect symmetrizing trace does not imply semisimplicity

Since $i^{-1}=-i$, the specialization of the quadratic relation in [F1] is $(T-i{)}^2=0$. The specialized basis is $\{1,T\}$, so $ϵ:=T-i$ is nonzero and $A\cong \mathbb{C}[ϵ]\mathrm{/}({ϵ}^2)$. The canonical trace has $\tau (1)=1$, $\tau (T)=0$, and $T^2=1+2iT$; therefore its Gram matrix in $(1,T)$ is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$. The algebra is commutative, so the trace is symmetric, and this matrix is invertible, so its trace pairing is perfect (also as guaranteed by [F2]).

As a left $A$-module, $Aϵ=\mathbb{C}ϵ$ is simple: $ϵ$ acts as zero on this one-dimensional complex vector space, and scalar elements of $\mathbb{C}$ act by scalar multiplication, so its only $A$-submodules are $0$ and $\mathbb{C}ϵ$ by [F4]. If $x=a+bϵ$ with $a\mathrm{\ne }0$, then $x$ is a unit, with inverse $a^{-1}-a^{-2}bϵ$; hence $Ax=A$, which is not simple because it contains the proper nonzero submodule $\mathbb{C}ϵ$. Thus a simple submodule cannot contain an element with nonzero constant coefficient. Every nonzero element with zero constant coefficient generates $\mathbb{C}ϵ$, so every simple submodule of $A$ is exactly $\mathbb{C}ϵ$.

The sum of all simple submodules of $A$ is consequently $\mathbb{C}ϵ$, which is proper because $1\mathrm{\notin }\mathbb{C}ϵ$. By [F3], the regular module is not a direct sum of simple submodules, so it is not semisimple. Step 1.1 gives a perfect symmetric trace on this finite-dimensional algebra, while this step proves its regular module is not semisimple. This refutes the claim. No choice principle is used.