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I know that the multiplicative affine group scheme $\mathbb{G}_m$ is diagonalizable, since the algebra that represents it is $k[X,X^{-1}]$, which is isomorphic to the group algebra $k[\mathbb{Z}]$. Note that I am using the definition of Waterhouse, "Introduction to Affine Group Schemes": an affine group scheme is diagonalizable iff it is represented by $k[M]$ for some abelian group $M$.

Does the fact that $\mathbb{G}_m$ is diagonalizable imply that for every representation $\mathbb{G}_m \to GL_n$ there are (compatible? (in what sense?)) bases for each $k-$algebra $R$ such that the image of $R^\times$ in $GL_n(R)$ consists of diagonal matrices wrt those bases?

Does this imply that every representation of $\mathbb{G}_m$ splits into $1-$dimensional representations? If not, how can I prove that it does split?

You can assume that $k$ is a field, if you need it, but I wouldn't mind a more general answer.

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This is not an answer but it's too long to be a comment, I'm just hoping it will shed a tiny bit of light.

I'm not an expert on affine group schemes at all, but here's what I can recollect about this thing.

Suppose $G=Spec(k[M])$ where $M$ is some abelian group and $X= Spec(R)$ is some affine scheme, $R$ a $k$-algebra, then a $G$-action on $X$ is essentially the same thing as an $M$-grading on $R$.

Indeed it corresponds to a map $R\to R\otimes k[M] = \displaystyle\bigoplus_{m\in M} R$, and the axioms for the group action tell you exactly that this map is a decomposition onto an $M$-grading (you have to get your hands into it to see what's happening - if you want I can add some details at this level, where I still understand things).

Now a representation $G\to GL_n$ is the same as an action of $G$ on $\mathbb{G}_a^n = Spec(k[X_1,...,X_n])$; so it corresponds to an $M$-grading of $k[X_1,...,X_n]$, and if you look at points on a specific $k$-algebra $A$ I think I remember a teacher of mine saying that this grading corresponds to an eigen-decomposition of the action of $G(A)$ on $\mathbb{G}_a^n(A) = A^n$, here I don't remember the details though (that's why this is not an answer, just a rather long comment).

Let me add that if $k$ is a field, a group-like element of $k[M]$ is an element of $M$ : indeed $\Delta (\sum_m a_m m) = \sum_m a_m m\otimes m$, so if it is group-like, then for all $m,n$, $a_ma_n = \delta_{m,n}a_m$ , therefore it must be $0$ or some element of $M$ : characters of $G$ correspond exactly to $M$, and again, if I recall correctly (but I'm fuzzy on the details) the eigenspaces of above correspond precisely to these characters of $G$.

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When $k$ is a field, the answer to your questions is yes, but I don't know how to prove it without using the general structure theory of linear algebraic groups. Let $\rho: \mathbb G_m \rightarrow \operatorname{GL}_n$ be the given representation. It suffices to replace $\mathbb G_m$ by $\operatorname{Im} \rho = \mathbb G_m/\operatorname{Ker} \phi$ and assume $\rho$ is a faithful representation: if $\rho$ is not the trivial homomorphism, then $\operatorname{Dim Ker} \rho = 0$, and so the image of $\rho$ is isomorphic to $\mathbb G_m$ (even though $\rho$ is not the identity on $\mathbb G_m$). This is because the $k$-rational image of a split torus is a split torus, and split tori are determined by their dimension.

So the question reduces to considering a $k$-closed subgroup $H$ of $\operatorname{GL}_n$ which is isomorphic to $\mathbb G_m$. In particular, $H$ is a split torus, and every split torus is contained in a maximal split torus. Every maximal split torus in $\operatorname{GL}_n$ is conjugate to the group of diagonal matrices by some matrix $g \in \operatorname{GL}_n(k)$.

So $\operatorname{Int} g$ defines an isomorphism of group schemes of $H$ onto a closed subgroup scheme of the group $T$ of diagonal matrices in $\operatorname{GL}_n$. This isomorphism will give you the splitting of $\rho$ into one-dimensional representations. Also, this isomorphism of group schemes is what will transfer each group $H(R) \cong R^{\ast}$ of $R$-rational points to a subgroup of diagonal matrices in $T(R) = (R^{\ast})^n$.

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