= 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 Coxeter system>[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> with
$$
T_i^2=-T_i.
$$
Let $w_0$ be the <Longest element of a finite Coxeter group>. Choose a simple generator $s$ in a component of rank at least two, put
$$
r=w_0sw_0,qquad v=w_0s,qquad X=T_v+T_{w_0}.
$$
The element $r$ is again a simple generator. The identities $rv=w_0$ and $\ell(w_0)=\ell(v)+1$ give
$$
T_rX=T_{w_0}-T_{w_0}=0.
$$
If $t\ne r$ is simple, then $t$ is a left descent of both $v$ and $w_0$: using $\ell(w_0u)=\ell(w_0)-\ell(u)$, one gets $\ell(tv)=\ell(v)-1$. Hence
$$
T_tX=-T_v-T_{w_0}=-X.
$$
The 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 $v=w_0s$ 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 $T_rX=0$, associativity now gives
$$
T_vX=T_{w_0}X=0,
\qquad
X^2=(T_v+T_{w_0})X=0.
$$
If $H$ 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 $W\cong(A_1)^m$, then
$$
H\cong k[T_1,\ldots,T_m]/(T_i(T_i+1))
\cong(k\times k)^{\otimes m},
$$
which is semisimple. This is the counterexample showing why the omitted rank condition is necessary.
Solved by gpt-5.6-sol high.
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