Representations of diagonalizable groups split into character eigenspaces

The correspondence holds for arbitrary $M$. Given a natural action $r$, evaluate it at the universal point $u={\mathrm{id}}_A\in G(A)$ and put $\rho (v)=r_A(u)(v\otimes 1)$. Naturality along $g:A\to R$ gives $r_R(g)(v\otimes 1)=({\mathrm{id}}_V\otimes g)\rho (v)$. At $g=\epsilon$ the identity action yields the counit identity. In $G(A\otimes A)$ the two points $p_1(a)=a\otimes 1$ and $p_2(a)=1\otimes a$ have product $\mathrm{\Delta }$; evaluating $r(p_1)r(p_2)=r(\mathrm{\Delta })$ at $v\otimes 1$ yields $(\rho \otimes \mathrm{id})\rho (v)=(\mathrm{id}\otimes \mathrm{\Delta })\rho (v)$. Conversely this coassociativity identity makes the displayed formula for $r_R(g)$ multiplicative; the counit gives the identity and $g^{-1}$ gives its inverse. These constructions are inverse by evaluation at $u$, and a subspace is stable under all $r_R(g)$ exactly when it is a subcomodule, by the same evaluation. No finite-type hypothesis or basis of $V$ is used. Finally, the group-like elements of $A$ are exactly $e_m$: if $a={\sum }_m{a}_m{e}_m$ is group-like, comparing coefficients in $\mathrm{\Delta }(a)=a\otimes a$ gives $a_m^2=a_m$ and $a_m{a}_n=0$ for $m\mathrm{\ne }n$, while $\epsilon (a)=1$ gives ${\sum }_m{a}_m=1$; over a field exactly one coefficient is $1$. Thus $X(G)=M$.

For $v\in V$ write $\rho (v)={\sum }_{m\in M}{v}_m\otimes e_m$ with $v_m\in V$, a finite sum by [F1]. Applying $(\mathrm{id}\otimes \mathrm{\Delta })$ and $(\rho \otimes \mathrm{id})$ to this expression and using coassociativity gives $(\rho \otimes \mathrm{id})\rho (v)={\sum }_m\rho (v_m)\otimes e_m$ and $(\mathrm{id}\otimes \mathrm{\Delta })\rho (v)={\sum }_m{v}_m\otimes e_m\otimes e_m$, so comparing the coefficients of the basis elements $e_n\otimes e_m$ of $k[M]\otimes k[M]$ in these two expressions gives $\rho (v_m)=v_m\otimes e_m$ for every $m$: indeed the coefficient of $e_n\otimes e_m$ with $n\mathrm{\ne }m$ vanishes on the right and equals the $e_n$-component of $\rho (v_m)$ on the left, and the remaining coefficient identifies the $e_m$-component of $\rho (v_m)$ with $v_m$.

The counit identity of [F3] applied to the expansion of [step 1.2] gives $v={\sum }_m{v}_m$ with each $v_m\in V_m=\{w:\rho (w)=w\otimes e_m\}$. Hence $V={\sum }_{m\in M}{V}_m$. If ${\sum }_m{w}_m=0$ is a finite relation with $w_m\in V_m$, applying $\rho$ gives ${\sum }_m{w}_m\otimes e_m=0$; extraction by ${\mathrm{id}}_V\otimes c_n$ gives $w_n=0$ for every $n$. Thus the sum is direct.

If $W\subseteq V$ is a subrepresentation, coefficient extraction in $\rho (W)\subseteq W\otimes k[M]$ gives $W={⨁}_m(W\cap V_m)$. An equivariant linear map preserves every weight, so a quotient has its corresponding weight decomposition; the middle term of any extension has the decomposition of step 2.1. No eigenspace is asserted to have dimension one. If $V$ is finite-dimensional, there are finitely many nonzero eigenspaces and choosing their finite bases expresses $V$ as a finite direct sum of character lines. Thus finite-dimensional representations are semisimple and $G$ is linearly reductive.

For an arbitrary-dimensional $V$, assume AC and choose a basis of each nonzero $V_m$ simultaneously by [F4]. Their union is a basis of $V$ by the direct sum decomposition, and its one-dimensional spans are character representations. This proves the additional arbitrary-dimensional character-line assertion, with AC spent only in these basis choices.