Groups of Multiplicative Type and Arithmetic Tori — Examples

The coordinate algebra is $k[t_1^{\pm 1},\dots ,t_r^{\pm 1}]=k[{\mathbb{Z}}^r]$. F1 shows that all characters are exactly its monomials; every exponent tuple occurs uniquely. These functions are defined over $k$, so the entire Galois action is trivial.

F1 identifies a map to ${\mathbf{G}}_m^s$ with a homomorphism ${\mathbb{Z}}^s\to {\mathbb{Z}}^r$, determined by the images of the $s$ basis vectors, giving the stated matrix and formulas. The module is torsion-free, so F2 identifies this group as a torus. The same formulas include rank zero and the unique map involving a trivial source or target as appropriate.

The displayed multiplication is multiplication of $x+y\sqrt{d}$; the norm $x^2-d{y}^2$ multiplies, so it preserves the equation. The identity is $(1,0)$ and the inverse is $(x,-y)$, and associative multiplication follows by direct expansion in the basis $1,\sqrt{d}$. This gives an affine group by F1. Over $L=k(\sqrt{d})$ put $t=x+\sqrt{d}y$. Then $t^{-1}=x-\sqrt{d}y$, and $x=(t+t^{-1})\mathrm{/}2$, $y=(t-t^{-1})\mathrm{/}(2\sqrt{d})$ give inverse algebra maps with $L[t,t^{-1}]$. They preserve multiplication, so $T_L\cong {\mathbf{G}}_m$.

The nontrivial $\sigma \in \mathrm{Gal}(L\mathrm{/}k)$ sends $t$ to $t^{-1}$, so it sends $t^n$ to $t^{-n}$. F2 gives the sign action on $\mathbb{Z}$; F3 makes $T$ a torus. It cannot be split: a split rank-one torus has trivial action, and no isomorphism of abelian groups $\mathbb{Z}\to \mathbb{Z}$ intertwines the sign action with the trivial action (its nonzero generator would have to equal its negative). For $d=-1$ over $\mathbb{R}$ the equation and product are exactly those of unit complex numbers.

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.