The top exterior power detects stabilizers of a subspace

Let $R$ be a $k$-algebra. The basis $e_1,\dots ,e_n$ of $V$ is an $R$-basis of $V_R$, and $W_R$ is the free direct summand $R{e}_1\oplus \cdots \oplus R{e}_d$ because $W$ is spanned by $e_1,\dots ,e_d$ and is complemented by the span of $e_{d+1},\dots ,e_n$. We work in the graded exterior algebra ${\bigwedge }_R(V_R)$ of [F2] and write ${\mathrm{\Lambda }}_R^k(V_R)$ for its degree-$k$ part; for $k\le n$ this is a free $R$-module, and the field case $R=k$ agrees with the exterior power of the statement by [F1], both having the wedges $e_I$ as basis.

Functoriality on ${\bigwedge }_R(V_R)$. An $R$-linear map $\alpha :V_R\to V_R$ induces the algebra map $T(\alpha )$ of the tensor algebra with $\alpha (v_1\otimes \cdots \otimes v_m)=\alpha v_1\otimes \cdots \otimes \alpha v_m$; it maps the defining ideal into itself because $v\otimes v\mapsto \alpha v\otimes \alpha v$ and the ideal is generated by these elements, so it descends to a graded $R$-linear map $\mathrm{\Lambda }(\alpha )$ with $\mathrm{\Lambda }(\alpha )(v_1\wedge \cdots \wedge v_m)=\alpha v_1\wedge \cdots \wedge \alpha v_m$, equal to ${\mathrm{\Lambda }}_R^m(\alpha )$ in degree $m$. If $\alpha$ is invertible with inverse $\beta$, then $\mathrm{\Lambda }(\alpha )\mathrm{\Lambda }(\beta )=\mathrm{\Lambda }(\beta )\mathrm{\Lambda }(\alpha )=\mathrm{id}$ on wedges and hence on the whole algebra by linearity, so each ${\mathrm{\Lambda }}_R^m(\alpha )$ is an $R$-linear automorphism; and for an $R$-submodule $U\subseteq V_R$ one has ${\mathrm{\Lambda }}_R^m(\alpha )({\mathrm{\Lambda }}_R^{m}U)={\mathrm{\Lambda }}_R^m(\alpha U)$, where ${\mathrm{\Lambda }}_R^{m}U$ is the span of $m$-fold wedges of elements of $U$. For $R=k$ this is [F5].

The increasing wedges span ${\mathrm{\Lambda }}_R^k(V_R)$. For $k\ge 1$ every element of ${\bigwedge }_R(V_R)$ is an $R$-linear combination of wedges of basis vectors, because $T_R(V_R)$ is spanned by pure tensors of basis vectors ([F2]) and the wedge is the quotient map. In the quotient, $u\wedge u=0$ for every $u\in V_R$, since $u\otimes u$ lies in the ideal of [F2]; applying this to $u+w$ and expanding bilinearly gives $u\wedge w=-w\wedge u$ for all $u,w$, so adjacent transpositions change a wedge by a sign. Consequently any wedge of basis vectors with two equal entries is zero: move the equal entries next to each other by adjacent transpositions, the wedge picks up a sign, and the adjacent product $u\wedge u$ vanishes. Every wedge of basis vectors with pairwise distinct indices therefore equals $\pm e_I$ for the corresponding increasing subset $I$, and the $e_I$ span ${\mathrm{\Lambda }}_R^k(V_R)$; for $k=0$ the element $1$ spans ${\mathrm{\Lambda }}_R^0(V_R)=R$.

The increasing wedges are $R$-independent. Fix a $k$-element subset $J=\{j_1<\cdots <j_k\}\subseteq \{1,\dots ,n\}$ and define, for $v_1,\dots ,v_k\in V_R$, the element ${\phi }_J(v_1,\dots ,v_k)\in R$ as the Leibniz determinant of the $k\times k$ matrix whose entry in row $m$ and column $\mathrm{\ell }$ is the coefficient of $e_{j_m}$ in $v_{\mathrm{\ell }}$; its columns depend $R$-linearly on $v_1,\dots ,v_k$ and equal columns occur when two of the vectors are equal, so by [F3] it is $R$-multilinear and alternating in $v_1,\dots ,v_k$. Extend ${\phi }_J$ $R$-linearly to a map ${\tilde{\psi }}_J:T_R(V_R)\to R$ on the basis of pure tensors of basis vectors of [F2], declaring it zero in tensor degree different from $k$. Then ${\tilde{\psi }}_J$ vanishes on every element $t(v\otimes v)t^{\mathrm{\prime }}$ with $t,t^{\mathrm{\prime }}$ pure tensors of basis vectors: expanding $v={\sum }_r{a}_r{e}_r$, such an element is a combination of pure tensors with two adjacent entries $e_r,e_s$ in the middle, and in degree $k$ it evaluates to ${\sum }_{r,s}{a}_r{a}_s{\phi }_J(\dots ,e_r,e_s,\dots )$, which is zero because a bilinear form alternating in two adjacent arguments satisfies ${\sum }_{r,s}{a}_r{a}_s\beta (e_r,e_s)=0$ over any commutative ring (the sum has ${\sum }_{r<s}{a}_r{a}_s(\beta (e_r,e_s)+\beta (e_s,e_r))+{\sum }_r{a}_r^2\beta (e_r,e_r)=0$). As these elements $R$-span the defining ideal of [F2], ${\tilde{\psi }}_J$ descends to an $R$-linear map ${\psi }_J:{\bigwedge }_R(V_R)\to R$. The matrix whose columns are $e_{i_1},\dots ,e_{i_k}$ in the rows $J$ is the identity matrix indexed by $J$ when $I=\{i_1<\cdots <i_k\}=J$, so ${\psi }_J(e_J)=1$ by [F3]; and when $I\mathrm{\ne }J$ some $\mathrm{\ell }$ has $i_{\mathrm{\ell }}\mathrm{\notin }J$ (as $\mathrm{\mid }I\mathrm{\mid }=\mathrm{\mid }J\mathrm{\mid }$), so column $\mathrm{\ell }$ is a zero column and ${\psi }_J(e_I)=0$ by [F3]. Therefore ${\sum }_I{c}_I{e}_I=0$ forces $c_J={\psi }_J(0)=0$ for every $J$, and the $e_I$ are $R$-independent.

Annihilator description of $W_R$. An element of $V_R$ is uniquely $v={\sum }_{i=1}^n{a}_i{e}_i$ with $a_i\in R$, and $w\wedge v={\sum }_{i=1}^n{a}_i{e}_1\wedge \cdots \wedge e_d\wedge e_i={\sum }_{i=d+1}^n{a}_i{e}_{\{1,\dots ,d,i\}}$, since wedges with a repeated index vanish by step 2.1. By step 2.2 the family of wedges $e_{\{1,\dots ,d,i\}}$, $d+1\le i\le n$, is part of the $R$-basis of ${\mathrm{\Lambda }}_R^{d+1}(V_R)$ and so is $R$-independent; hence $w\wedge v=0$ if and only if $a_i=0$ for all $i>d$, that is, if and only if $v\in W_R$.

If $\alpha W_R=W_R$ then $({\mathrm{\Lambda }}^d\alpha )(D_R)=D_R$. The submodule ${\mathrm{\Lambda }}_R^d(W_R)$ of ${\mathrm{\Lambda }}_R^d(V_R)$ is generated by $w$: any $d$-fold wedge of elements of $W_R$ expands in the $R$-basis $e_1,\dots ,e_d$ of $W_R$ into a combination of wedges with repeated indices and of the single wedge $w$. Hence $D_R={\mathrm{\Lambda }}_R^d(W_R)$, and step 1.2 gives $({\mathrm{\Lambda }}_R^d\alpha )(D_R)={\mathrm{\Lambda }}_R^d(\alpha W_R)={\mathrm{\Lambda }}_R^d(W_R)=D_R$.

If $({\mathrm{\Lambda }}^d\alpha )(D_R)=D_R$ then $\alpha W_R=W_R$. Since $D_R=Rw$ and ${\mathrm{\Lambda }}_R^d\alpha$ is an automorphism of ${\mathrm{\Lambda }}_R^d(V_R)$ preserving $D_R$ (step 1.2), its restriction $D_R\to D_R$ is bijective and therefore ${\mathrm{\Lambda }}_R^d\alpha (w)=cw$ for a unique unit $c\in R^{\times }$. Let $v\in W_R$; by step 3.1 $w\wedge v=0$, so $0={\mathrm{\Lambda }}_R^{d+1}\alpha (w\wedge v)=({\mathrm{\Lambda }}_R^d\alpha )(w)\wedge {\mathrm{\Lambda }}_R\alpha (v)=c(w\wedge \alpha v)$, and since $c$ is a unit, $w\wedge \alpha v=0$, so $\alpha v\in W_R$ by step 3.1. Thus $\alpha W_R\subseteq W_R$. The same argument applied to ${\alpha }^{-1}$, whose induced map $({\mathrm{\Lambda }}_R^d\alpha {)}^{-1}$ also preserves $D_R$, gives ${\alpha }^{-1}{W}_R\subseteq W_R$, hence $W_R\subseteq \alpha W_R$, and therefore $\alpha W_R=W_R$.

The stabilizers agree. Let $(V,r)$ be a rational representation of an affine group scheme $G$ over $k$ and let $g\in G(R)$ for a $k$-algebra $R$; the $R$-point of the representation is the $R$-linear automorphism $r_R(g)$ of $V_R$, and by [F6] the induced action on ${\mathrm{\Lambda }}^{d}V$ is the rational action ${\mathrm{\Lambda }}_R^d(r_R(g))$. Steps 3.2 and 4.1 give $r_R(g)W_R=W_R$ if and only if $({\mathrm{\Lambda }}_R^d(r_R(g)))(D_R)=D_R$, that is, the $R$-points of the stabilizer of the subspace $W$ and of the stabilizer of the line $D$ coincide for every $k$-algebra $R$.

By [F4] the scheme-theoretic stabilizers of $W$ and of the line $D$ are the closed subgroup schemes of $G$ defined by exactly these two functors of points; a closed subscheme of an affine scheme is determined by its functor of points, so the two closed subgroup schemes are equal. This completes both the base-change equivalence for every $k$-algebra $R$ and the scheme-theoretic stabilizer statement.