A degenerate one-dimensional Clifford algebra need not be semisimple

Take $V=ke$ with $B(e,e)=Q(e)=0$ and put $\epsilon =\bar{e}$. By [F1] and [F2], $A=\mathrm{C}\mathrm{l}(V,0)=k1\oplus k\epsilon$, ${\epsilon }^2=0$, and $\epsilon \mathrm{\ne }0$. For $a,b,c\in k$, $(a1+b\epsilon )(c\epsilon )=ac\epsilon$, so [F3] shows $k\epsilon$ is a nonzero proper left ideal and hence a submodule of the left regular module. Therefore that module is not simple.

If the two-dimensional regular module were a direct sum of simple modules, the summands' positive $k$-dimensions would force either one two-dimensional simple summand or two one-dimensional simple summands. The first case is excluded by step 1.1.

On any one-dimensional left $A$-module, the action of $\epsilon$ is multiplication by some $\lambda \in k$; ${\epsilon }^2=0$ gives ${\lambda }^2=0$, hence $\lambda =0$ because $k$ is a field. Thus in the second case $\epsilon$ acts as zero on the direct sum, while in the regular module $\epsilon \cdot 1=\epsilon \mathrm{\ne }0$. This contradiction proves the regular module is not semisimple.