 Solution

There is a missing hypothesis in the printed claim: it is false when every irreducible component of $W$ has type $A_1$. The intended statement holds as soon as $W$ has an irreducible component of rank at least two, which we now assume.

Since $q=p$, every Hecke parameter of a BN-pair vanishes in $k$, and $H$ is the 0-Hecke algebra withLet $w_0$ be the Longest element of a finite Coxeter group. Choose a simple generator $s$ in a component of rank at least two, putThe element $r$ is again a simple generator. The identities $rv=w_0$ and $\ell (w_0)=\ell (v)+1$ giveIf $t\ne r$ is simple, then $t$ is a left descent of both $v$ and $w_0$: using $\ell (w_{0}u)=\ell (w_0)-\ell (u)$, one gets $\ell (tv)=\ell (v)-1$. HenceThe one-dimensional subspace $kX$ is therefore a left ideal. It is nonzero because $T_v$ and $T_{w_0}$ are distinct basis elements.

Every reduced expression in a Coxeter group for $v=w_{0}s$ contains $r$: in an irreducible finite component of rank at least two, deleting one final generator from $w_0$ does not remove any vertex from its support. A reduced expression for $w_0$ contains $r$ as well. Since $T_{r}X=0$, associativity now givesIf $H$ were a semisimple algebra, the left ideal $kX$ would be a direct summand of the regular module. The corresponding projection would produce a nonzero idempotent in $kX$, impossible because $(kX{)}^2=0$. Thus $H$ is not semisimple.

For completeness, if $W\cong (A_1{)}^m$, thenwhich is semisimple. This is the counterexample showing why the omitted rank condition is necessary.

Solved by gpt-5.6-sol high.