Universal Clifford Algebras and Their PBW Basis: Examples and Counterexamples

Since $(e)$ is an ordered basis and $Q(e)=B(e,e)=a$, [F1] gives the basis $1,\bar{e}$ and [F2] gives ${\bar{e}}^2=a1$.

Bilinearity and the unit law determine every product of basis elements from this relation; when $a=0$, [F1] still ensures $\bar{e}\mathrm{\ne }0$, while [F2] gives ${\bar{e}}^2=0$.

Since $Q(v)=0$ for every $v$, [F1] and [F2] quotient $T_k(V)$ by the same two-sided ideal. The map $[t{]}_{\mathrm{C}\mathrm{l}}\mapsto [t{]}_{\mathrm{\Lambda }}$ is therefore well defined, preserves the induced algebra operations, and has the inverse $[t{]}_{\mathrm{\Lambda }}\mapsto [t{]}_{\mathrm{C}\mathrm{l}}$.

By [F3] and [F4], the same ordered square-free monomials are bases on the two sides; for $n=0$ both lists consist only of $1$. Thus the canonical presentation isomorphism has the asserted basis description, including the degenerate case.

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.