Let $k$ be a field of characteristic zero. The general linear group $\mathrm{GL}(n,k)$ has a subgroup $\mathrm{D}(n,k)$ consisting of invertible diagonal matrices. These are linear algebraic groups and the inclusion is a map of linear algebraic groups. So, we can restrict any algebraic representation of the former group to one of the latter, giving us a functor

$$ r_k \colon \mathsf{Rep}(\mathrm{GL}(n,k)) \to \mathsf{Rep}(\mathrm{D}(n,k)) $$

I'm trying to do this:

Find a "simple" proof that this functor is essentially injective: i.e., if two representations of $\mathrm{GL}(n,k)$ become isomorphic when restricted to $D(n,k)$, they had to be isomorphic in the first place.

Of course different people have ideas on what counts as "simple". By a "simple" proof I mean one that does not rely on developing the representation theory of reductive algebraic groups, or on classifying representations of $\mathrm{GL}(n,k)$ using Young diagrams and Schur-Weyl duality.

I started trying to take a proof based on the theory of reductive algebraic groups and distill it to the minimum. I got an argument that still seems a bit roundabout and unsatisfying. It goes like this.

First consider the case of an algebraically closed field $K$ of characteristic zero. Show that:

• $\mathrm{GL}(n,K)$ is a reductive algebraic group.
• Any finite-dimensional algebraic representation of a reductive algebraic group is determined up to isomorphism by its character.
• The semisimple elements of $\mathrm{GL}(n,K)$ form a Zariski dense set, so the character of any algebraic representation is determined by its restriction to the semisimple elements. (Simple enough.)
• The character of any finite-dimensional representation is a class function, i.e. invariant under conjugation. (Immediate.)
• When $K$ is algebraically closed, every semisimple element of $\mathrm{GL}(n,K)$ is conjugate to a diagonal matrix. (Simple enough.)
• Thus, every finite-dimensional algebraic representation of $\mathrm{GL}(n,K)$ is determined up to isomorphism by the restriction of its character to $\mathrm{D}(n,K)$. (Immediate.)
• A fortiori, every finite-dimensional algebraic representation of $\mathrm{GL}(n,K)$ is determined up to isomorphism by its restriction to $\mathrm{D}(n,K)$. (Immediate.)

Then, if $k$ is of characteristic zero but not algebraically closed, let $K$ be an algebraic closure of $k$. I get a square of functors that commutes up to natural isomorphism, where the top row is this:

$$ \mathsf{Rep}(\mathrm{GL}(n,k)) \xrightarrow{r_k} \mathsf{Rep}(\mathrm{D}(n,k)) $$

the bottom row is this:

$$ \mathsf{Rep}(\mathrm{GL}(n,K)) \xrightarrow{r_K} \mathsf{Rep}(\mathrm{D}(n,K)) $$

and the vertical arrows are "tensoring with $K$", i.e. $K \otimes_k -$. The vertical arrows and the bottom arrow ($r_K$) are essentially injective. Thus, going down and then across is essentially injective. Thus, going across and then down is essentially injective. But the composite of two functors can only be essentially injective if the functor you do first is essentially injective. Thus, the top arrow ($r_k$) is essentially injective.

Most of this argument seems "simple enough" to me, but I'm not happy with these steps:

• $\mathrm{GL}(n,k)$ is a reductive algebraic group.

• Any finite-dimensional algebraic representation of a reductive algebraic group is determined up to isomorphism by its character.

because we are proving a fact for general reductive algebraic groups that might be easier to prove for $\mathrm{GL}(n,k)$. So, one solution to my problem would be to solve this problem:

Find a "simple" proof that any finite-dimensional algebraic representation of $\mathrm{GL}(n,k)$ is determined up to isomorphism by its character.

But there might be some other way.