The multiplicative group scheme mu p is not a smooth torus

F1 gives $G=D_k(\mathbb{Z}\mathrm{/}p\mathbb{Z})$ and $X^{\ast }(G)=\mathbb{Z}\mathrm{/}p\mathbb{Z}$ with trivial Galois action. Thus it is of multiplicative type by F2 and is not a torus by F3. Its algebra is $k[u]\mathrm{/}(u^p)$ with $u=t-1$, so it has a nonzero nilpotent. For any field extension $K\mathrm{/}k$, $G(K)=\{1\}$, although its coordinate ring has dimension $p$ over $k$.

Its local ring $k[u]\mathrm{/}(u^p)$ has only one prime ideal, $(u)$, so has Krull dimension zero. The quotient $(u)\mathrm{/}(u^2)$ has dimension one over its residue field $k$, since $p\ge 2$. Thus this Noetherian local ring is not regular by F4. Already over $k$ the fibre fails regularity, so it is not geometrically regular and $G\to \mathrm{Spec}k$ is not smooth by F4. Together with step 1.1, this refutes the statement.