Every normal subgroup of an affine group is a representation kernel

First let $k$ be algebraically closed. Choose by [F1] a representation $V$ and a line $L$ with stabilizer $N$. The action of $N$ on $L$ is a character $\chi$. By [F2], some representation $V^{\mathrm{\prime }}$ contains a nonzero $N$-weight subspace $D$ of character ${\chi }^{-m}$. The subspace $T=L^{\otimes m}\otimes D\subset V^{\otimes m}\otimes V^{\mathrm{\prime }}$ has stabilizer exactly $N$. To see this on any algebra $R$, suppose an automorphism in each factor preserves $W_R\otimes D_R$, with $W,D$ nonzero subspaces. For a transformed basis vector $u$ of $W_R$ and a transformed basis vector $d$ of $D_R$, project $u\otimes d$ to $(V_R\mathrm{/}{W}_R)\otimes V_R^{\mathrm{\prime }}$. It is zero. The vector $d$ is unimodular because it is part of a transformed basis, so contracting with a functional taking $d$ to $1$ gives $u<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>W_R=0$. Interchanging the factors gives preservation of $D_R$, and applying the inverse gives equality of both submodules. Thus the tensor-subspace stabilizer is the intersection of the factor stabilizers. Likewise, the tensor line $L_R^{\otimes m}$ determines $L_R$: for a unimodular generator $u$ of a transformed line choose a functional with value $1$ on $u$, and contract $u^{\otimes m}$ in all but one slot after projecting the remaining slot to $V_R\mathrm{/}{L}_R$. Thus its stabilizer is the stabilizer of $L_R$. Since $N$ stabilizes $D$, the claimed intersection is exactly $N$. The action of $N$ on $T$ is trivial, since the characters cancel.

Taking the top exterior power of $T$ and of its ambient representation gives a line $L_0$ with stabilizer $N$, by the wedge calculation in [F1], and $N$ acts trivially on $L_0$. Write $B$ for this ambient representation and $B^N$ for the kernel of the linear map $b\mapsto {\rho }_N(b)-b\otimes 1$ into $B\otimes k[N]$. Tensoring this kernel with every $k$-algebra $R$ preserves it, since all $k$-modules are flat. Thus $B^N\otimes R$ is exactly the vectors fixed by every $N$-point after every further algebra extension: the universal point of $N$ tests the coaction equality. Normality makes this space $G$-stable. Indeed, for $g\in G(R)$, $v\in B^N\otimes R$ and $n\in N(R^{\mathrm{\prime }})$, where $R^{\mathrm{\prime }}$ is any $R$-algebra, $n(gv)=g(g^{-1}ng)v=gv$. The representation on $B^N$ has kernel containing $N$. Its kernel fixes $L_0\subset B^N$, hence is contained in the line stabilizer $N$. These inclusions hold on all algebras, so its scheme-theoretic kernel equals $N$.

For arbitrary $k$, extend to an algebraic closure $\bar{k}$ by [F3] and apply steps 1.1–2.1 there. The resulting representation is given by finitely many matrix coefficients in $\bar{k}{\otimes }_{k}k[G]$, its inverse determinant and the finitely many relations expressing its group identities. All coefficient scalars lie in a finite subextension $K\mathrm{/}k$. Equality of its kernel ideal with $I_N\otimes \bar{k}$ can also be descended to a finite such extension: the representation-kernel ideal is generated by its matrix coefficients minus those of the identity; $I_N$ has a finite generating set by [F3]; expressing each set of generators in terms of the other uses only finitely many additional scalars. Enlarge $K$ to contain them. The representation on $K^n$ then exists over $K$, its group identities hold by injectivity of $k[G]\otimes K\to k[G]\otimes \bar{k}$ and its tensor square, and its kernel is exactly $N_K$. This argument permits inseparable $K\mathrm{/}k$.

Let $E$ be the underlying $k$-vector space of $K^n$. For every $k$-algebra $R$, extend $g\in G(R)$ to $G(K{\otimes }_{k}R)$ and apply the $K$-representation from step 3.1. This gives an invertible $R$-linear map on $E{\otimes }_{k}R$ and hence a representation of $G$ on $E$: choosing a $k$-basis of $K$ expresses its entries as regular $k[G]$-functions, and multiplication and inversion follow from the $K$-representation. This automorphism is the identity precisely when $g$ extended to $K\otimes R$ lies in $N(K\otimes R)$. Since $R\to K\otimes R$ is faithfully flat, it is injective, and the vanishing of every generator of $I_N$ after this extension is equivalent to its vanishing in $R$. Thus the kernel on $R$-points is $N(R)$ for every $R$, proving the scheme assertion. AC is used through [F2] and [F3].