Tensor products, exterior powers and Hom spaces of finite-dimensional rational representations are rational

The right-hand side of (a) is $k$-bilinear in $(v,w)$, since both comodule maps are $k$-linear and scalars move between tensor factors over $k$; [F3] therefore gives a unique additive group homomorphism $c:V{\otimes }_{k}W\to V{\otimes }_{k}W{\otimes }_{k}A$. This homomorphism is $k$-linear: on every decomposable tensor, $c(\lambda (v\otimes w))=c((\lambda v)\otimes w)=\lambda c(v\otimes w)$ by the formula, and decomposable tensors generate the source additively.

Counit axiom. Applying $\mathrm{id}\otimes \epsilon$ to $c(v\otimes w)=\sum v_i\otimes w_j\otimes a_i{b}_j$ and using that $\epsilon$ is multiplicative and $(\mathrm{id}\otimes \epsilon ){\rho }_V={\mathrm{id}}_V$, $(\mathrm{id}\otimes \epsilon ){\rho }_W={\mathrm{id}}_W$ gives $\sum v_i\otimes w_j\epsilon (a_i)\epsilon (b_j)=v\otimes w$, so $(\mathrm{id}\otimes \epsilon )c={\mathrm{id}}_{V\otimes W}$.

Coassociativity. Write the comultiplications in Sweedler notation, ${\rho }_V(v)=\sum v_{(1)}\otimes v_{(2)}$, ${\rho }_W(w)=\sum w_{(1)}\otimes w_{(2)}$. Then $(c\otimes \mathrm{id})c(v\otimes w)=\sum v_{(1)(1)}\otimes w_{(1)(1)}\otimes v_{(1)(2)}{w}_{(1)(2)}\otimes v_{(2)}{w}_{(2)}$, while $(\mathrm{id}\otimes \mathrm{\Delta })c(v\otimes w)=\sum v_{(1)}\otimes w_{(1)}\otimes v_{(2)(1)}{w}_{(2)(1)}\otimes v_{(2)(2)}{w}_{(2)(2)}$; the two sums agree after rewriting the first three tensor factors with the coassociativity identities for ${\rho }_V$ and ${\rho }_W$ and using multiplicativity of $\mathrm{\Delta }$ and commutativity of $A$ in the last two factors. Hence $(c\otimes \mathrm{id})c=(\mathrm{id}\otimes \mathrm{\Delta })c$.

The associated representation. By [F1] the representation associated with $c$ acts on $R$-points by sending $v\otimes w$ to $(\mathrm{id}\otimes {\mathrm{ev}}_g)c(v\otimes w)=\sum v_i\otimes w_j{a}_i(g)b_j(g)$, and this equals $({\sum }_i{v}_i{a}_i(g))\otimes ({\sum }_j{w}_j{b}_j(g))=(g\cdot v)\otimes (g\cdot w)$ because the two $R$-valued sums are exactly the actions of $g$ on $v$ and on $w$ under [F1]. This proves (a).

Hom is rational. For a $k$-algebra $R$ and $g\in G(R)$, the tensor product of the rational representations $V^{\ast }$ and $W$ acts on $V^{\ast }{\otimes }_{k}W$ by $g\cdot (\xi \otimes w)=(g\cdot \xi )\otimes (g\cdot w)$ by step 2.3 applied to the pair $(V^{\ast },W)$, and $(g\cdot \xi )(v)=\xi (g^{-1}v)$ by [F4]. Under the isomorphism $V^{\ast }{\otimes }_{k}W\to {\mathrm{Hom}}_k(V,W)$ of [F5], the element $\xi \otimes w$ corresponds to the map $f(v)=\xi (v)w$, and $g\cdot (\xi \otimes w)$ corresponds to $v\mapsto \xi (g^{-1}v)g\cdot w=g\cdot (\xi (g^{-1}v)w)=g\cdot f(g^{-1}v)$; the transport is therefore the action $(g\cdot f)(v)=g\cdot f(g^{-1}v)$ on ${\mathrm{Hom}}_k(V,W)$, which is thus a rational representation isomorphic to $V^{\ast }{\otimes }_{k}W$. This proves (b).

Exterior powers are rational and commute with scalar extension. For $d=0$, the exterior power is $k$ with the trivial action, and its base change is $R$ with the identity map. For $d\ge 1$, write ${\rho }_V(v)={\sum }_i{v}_i\otimes a_i$ and define $c_d:{\mathrm{\Lambda }}_k^{d}V\to {\mathrm{\Lambda }}_k^{d}V{\otimes }_{k}A$ by $c_d(v_1\wedge \cdots \wedge v_d)={\sum }_{i_1,\dots ,i_d}{v}_{1,i_1}\wedge \cdots \wedge v_{d,i_d}\otimes a_{1,i_1}\cdots a_{d,i_d}$, where ${\rho }_V(v_j)={\sum }_i{v}_{j,i}\otimes a_{j,i}$. This formula is alternating in $v_1,\dots ,v_d$: if two inputs coincide, terms with distinct corresponding indices cancel in pairs by $x\wedge y=-y\wedge x$ and commutativity of $A$, and equal-index terms vanish; hence [F9] makes it well-defined. The counit and coassociativity axioms follow from multiplicativity of $\epsilon$, coassociativity of ${\rho }_V$, and multiplicativity of $\mathrm{\Delta }$, as in steps 2.1--2.2. Thus ${\mathrm{\Lambda }}_k^{d}V$ is a comodule, and its associated action sends each decomposable wedge to the wedge of the actions by [F1] and the computation of step 2.3 with $d$ factors. Now choose an ordered basis $e_1,\dots ,e_n$ of $V$ and write $e_I$ for its increasing-index wedges. For $1\le d\le n$, [F8] gives that the $e_I$ form a $k$-basis of ${\mathrm{\Lambda }}_k^{d}V$; if $d>n$, expanding decomposable wedges in the $e_i$ gives only repeated-index wedges, which vanish in the quotient defining ${\mathrm{\Lambda }}_k^{d}V$, so that space is zero. By [F6], the $e_i\otimes 1$ form an $R$-basis of $V_R$, and [F7] gives the matching $R$-basis $e_I^R$ of ${\mathrm{\Lambda }}_R^d(V_R)$ (or zero for $d>n$). The alternating $k$-multilinear map $(v_1,\dots ,v_d)\mapsto (v_1\otimes 1)\wedge \cdots \wedge (v_d\otimes 1)$ induces a map ${\mathrm{\Lambda }}_k^{d}V\to {\mathrm{\Lambda }}_R^d(V_R)$ by [F9]; multiplying its values by $r\in R$ gives a $k$-balanced map ${\mathrm{\Lambda }}_k^{d}V\times R\to {\mathrm{\Lambda }}_R^d(V_R)$, so [F3] induces a group homomorphism ${\beta }_d:({\mathrm{\Lambda }}_k^{d}V){\otimes }_{k}R\to {\mathrm{\Lambda }}_R^d(V_R)$. It is $R$-linear because ${\beta }_d(x\otimes ar)=a{\beta }_d(x\otimes r)$ on elementary tensors, which generate additively. It sends $e_I\otimes 1$ to $e_I^R$, hence is an isomorphism by the two basis descriptions, with both sides zero when $d>n$. For every $g\in G(R)$, the tensor-power map of $r_R(g)$ preserves the ideal generated by the squares $u\otimes u$ in the exterior algebra of [F7], so descends to ${\mathrm{\Lambda }}_R^d(r_R(g))$; both it and the base-changed action send $e_I\otimes 1$ to $r_R(g)(e_{i_1}\otimes 1)\wedge \cdots \wedge r_R(g)(e_{i_d}\otimes 1)$. They therefore agree on the basis and under ${\beta }_d$. This proves (c).

The degree-zero and positive-degree cases above establish the stated exterior-power action and its base change for every $d\ge 0$.