Every affine-group comodule is a union of finite-dimensional rational submodules

Write $c(v)={\sum }_{i=1}^n{v}_i\otimes h_i$ with the $h_i$ linearly independent, eliminating redundancies from a finite tensor expression, and put $W_v=\mathrm{span}(v_1,\dots ,v_n)$. The counit gives $v={\sum }_i\epsilon (h_i)v_i\in W_v$. With $q:V\to V\mathrm{/}{W}_v$, coassociativity gives ${\sum }_{i}c(v_i)\otimes h_i={\sum }_i{v}_i\otimes \mathrm{\Delta }(h_i)$. Apply $q$ to the first factor: the right side is zero and independence of the $h_i$ gives $(q\otimes \mathrm{id})c(v_i)=0$. Thus $c(v_i)\in W_v\otimes H$, since the kernel of $q\otimes \mathrm{id}$ is $W_v\otimes H$ over a field. Indeed write any finite tensor expression with independent second-factor coefficients; its image under $q\otimes \mathrm{id}$ is zero exactly when each first-factor coefficient lies in $W_v$. This proves the kernel assertion without an infinite basis.

Sums of finitely many $W_v$ are finite-dimensional subcomodules, so they contain any prescribed finite subset; sums of two such subcomodules also show directedness. Evaluating coassociativity at $g,h$ gives $r(g)r(h)=r(gh)$, and evaluating the counit gives $r(e)=\mathrm{id}$, so $r(g^{-1})$ is the inverse of $r(g)$. Choose a finite basis of $W$ and write $c(w_j)={\sum }_i{w}_i\otimes a_{ij}$. The matrix entries $a_{ij}$ are regular functions on $G$, and its determinant is invertible at each point. Its inverse determinant is regular: the inverse matrix is the regular matrix $(S(a_{ij}))$, so determinants multiply to 1 as functions in $H$. Therefore $g\mapsto (a_{ij}(g))$ is a morphism to $GL(W)$, proving rationality. Only finite bases and finite elimination were used.