Group Schemes of Finite Type over a Field — Examples

For ${\mathbf{G}}_a$, the comorphisms of multiplication, identity and inverse send $x$ respectively to $x\otimes 1+1\otimes x$, $0$, and $-x$. They are algebra maps and hence morphisms by [F2]. Evaluation on $R$ identifies its points with $R$ and its operations with addition, zero, and negation. For ${\mathbf{G}}_m$, the corresponding formulas are $t\mapsto t\otimes t$, $t\mapsto 1$, and $t\mapsto t^{-1}$. Each image of $t$ is a unit, so these maps are defined on the Laurent algebra. Evaluation identifies its points with $R^{\times }$ and its operations with multiplication, one and inversion. These formulas satisfy the group-object identities as ring identities and hence as scheme morphisms by [F1]–[F2].

For $X=(x_{ij})$, define multiplication by $x_{ij}\mapsto {\sum }_l{x}_{il}\otimes x_{lj}$. Its determinant is $(d\otimes 1)(1\otimes d)$ by [F3], a unit, so the formula extends to the localized coordinate ring. The identity has $x_{ij}\mapsto {\delta }_{ij}$, with determinant one. Define inversion by the entries of $d^{-1}\mathrm{adj}(X)$; they belong to the same localized algebra. Its determinant is a unit, since the adjugate identity gives $X{X}^{-1}=I$ and determinant multiplicativity gives $\det (X^{-1})=d^{-1}$. Thus inversion also gives a morphism. The points of the localized spectrum are exactly matrices with unit determinant, equivalently invertible matrices by [F3].

Matrix associativity, the identity matrix, and the two inverse identities in [F3] verify all group identities on ${\mathrm{GL}}_n(R)$, for every $R$. They also verify the scheme identities: each domain in those identities is affine by [F2]; testing its coordinate algebra with its universal point tests the morphisms themselves. All three displayed coordinate algebras are finitely generated over $k$ (write the determinant inverse as a generator subject to $zd-1=0$), so the schemes are finite type. They are therefore group schemes by [F1]. No field-valued-point or smoothness argument substitutes for these formulas.

For every commutative $k$-algebra $R$, ${\alpha }_p(R)=\{a\in R:a^p=0\}$ is an additive subgroup of $R$: $(a+b{)}^p=a^p+b^p$ and $(-a{)}^p=(-1{)}^p{a}^p$. Also ${\mu }_p(R)=\{u\in R^{\times }:u^p=1\}$ is a multiplicative subgroup of $R^{\times }$. The closed immersions into ${\mathbf{G}}_a$ and ${\mathbf{G}}_m$ therefore give the induced group laws by [F1]–[F2]. AC is carried through the criterion in [F1], as recorded in [F4]. Each coordinate algebra has dimension $p$ over $k$, so is finite type. The formula $t=1+x$ identifies their underlying rings because $(1+x{)}^p-1=x^p$. In a field $a^p=0$ forces $a=0$, and $t^p=1$ forces $(t-1{)}^p=0$, hence $t=1$. Thus both rational-point groups are singleton while both schemes retain a nonzero nilpotent coordinate.

Every group-scheme homomorphism $f:{\alpha }_p\to {\mathbf{G}}_m$ corresponds by [F3] to a unit $g(x)={\sum }_{j=0}^{p-1}{c}_j{x}^j$ in $k[x]\mathrm{/}(x^p)$ satisfying $g(0)=1$ and $g(x+y)=g(x)g(y)$ in $k[x,y]\mathrm{/}(x^p,y^p)$. Compare coefficients of $x^{r-1}y$ for $1\le r<p$: the left side has coefficient $r{c}_r$, the right side $c_{r-1}{c}_1$. Since $c_0=1$ and $1,\dots ,p-1$ are invertible in $k$, induction gives $c_r=c_1^r\mathrm{/}r!$. Now compare the coefficient of $x^{p-1}y$: the left side is zero because every term of $g(x+y)$ has total degree less than $p$, while the right side is $c_{p-1}{c}_1=c_1^p\mathrm{/}(p-1)!$. Thus $c_1=0$ and all $c_r=0$ for $r>0$. This also covers $p=2$. Hence every such homomorphism is the trivial one, $g=1$.

If ${\alpha }_p\cong {\mu }_p$ as group schemes, compose that isomorphism with the closed subgroup inclusion ${\mu }_p\hookrightarrow {\mathbf{G}}_m$. The result would be nontrivial: on coordinate rings the inclusion pulls $t$ back to its nonconstant class in $k[t]\mathrm{/}(t^p-1)$, and an isomorphism cannot send $t-1\mathrm{\ne }0$ to zero. This contradicts step 2.1. Thus the two group schemes are not isomorphic despite their isomorphic underlying schemes and rational-point groups. Nilpotent test algebras distinguish their laws; for example their common coordinate $x=t-1$ has the additional product term $x\otimes x$ for ${\mu }_p$ written in the counterexample.