Defining a torus $T$ a certain way, prove that $T\cong (k^*)^n$.

82 Views Asked by At

The Question:

Suppose we are given the definition that a torus $T$ of a linear algebraic group over $k$ is a subgroup isomorphic to $\Bbb D_n$, show that $$T\cong(k^*)^m$$ for some $m\in\Bbb N$.

NB: Here $\Bbb D_n$ is the group of nonsingular, diagonal $n\times n$ matrices with entries in $k$.

Context:

I have looked in Springer, Borel, and Humphreys; a proof is nowhere obvious in them. The text "Linear Algebraic Groups", by Tom De Medts doesn't prove it (as far as I can tell). Herzig's text of the same name defines the theorem into existence. According to Approach0 and this MSE search, this question is new to MSE.

Why am I asking?

Because I'm studying for a postgraduate research degree in linear algebraic groups. I am writing a report, and part of the process of doing so entails revision of the basic material.

Thoughts:

My intuition is that I "simply" need to argue in terms of the dimension of $\Bbb D_n$.

Wouldn't it be $T\cong \Bbb D_n\cong (k^*)^{n-1}$, since $\Bbb D_n$ is the group of nonsingular diagonal matrices? (The $n$th entry is determined by the first $(n-1)$th entries along the diagonal.)


Please help :)