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Let $k$ be a field and $G$ be an affine algebraic group. We assume that $G$ is a $k$-subvariety of $\operatorname{GL}(V) \subset \operatorname{End}(V)$ for appropriate vector space $V$ of dimension $d= \dim{V}$. Let $k[G]$ be the coordinate ring of $G$.

Why and how is it possible to associate to every element $P \in k[G]$ a degree $\operatorname{deg}(P)$?

Although the assomption $G \subset \operatorname{End}(V)$ implies that $k[G]= k[X^iY^j \mathrel\vert 1 \le i,j \le d]/I$ is a quotient of a polynomial ring with ideal $I=I(G)$, there is no reason why it should be possible to associate a degree to an element from $k[G]$.

But seemingly that's exactly what is done in the proof of Proposition 4.6 (Chevalley) from the notes Réseaux des groupes de Lie by Yves Benoist (page 38).

Note the script is writen in French and seemingly the notation for degree of a $P \in k[G]$ is $d^{\circ} P$. But as I explained above this definition make no sense (at least) to me as long as $k[G]$ is not isomorphic to a polynomial ring.

Does somebody know what Benoist has in mind there?

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    $\begingroup$ Could one simply declare the degree of $P$ to be the degree of a minimal-degree representative in $k[\operatorname{End}(V)]$? It seems that all that is needed in the proof of Proposition 4.6 is a filtration of $I(\mathbf H)$. $\endgroup$
    – LSpice
    Commented Nov 14, 2021 at 1:53
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    $\begingroup$ The main point is that any finite-dimensional subspace of a representation of a linear algebraic group is contained in a finite-dimensional representation. (The same trick is used to prove that every linear algebraic group embeds faithfully in some $\operatorname{GL}_n$.) This is proven in a more Hopf-algebraic way in, for example, Proposition 1.9 of Borel's Linear algebraic groups. $\endgroup$
    – LSpice
    Commented Nov 14, 2021 at 1:58

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