3
$\begingroup$

The following Question has a yes answer when $K$ is algebraically closed. I am looking for an elementary proof of it, as well as an answer for arbitrary $K$ of characteristic $0$.

Question. Is a connected subgroup of $(K^*,\times)^n$ defined by polynomial equations and having Zariski dimension 1 isomorphic to $K^\times$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Over any field, every smooth connected algebraic subgroup of $\mathbb{G}_m^n$ of dimension one is isomorphic to $\mathbb{G}_m$. First note that such a subgroup is diagonalizable because it has a faithful diagonalizable representation. Then use that there is a contravariant category equivalence between diagonalizable groups and finitely generated $\mathbb{Z}$-modules. The statement then becomes that every torsion-free quotient of $\mathbb{Z}^n$ of rank one is isomorphic to $\mathbb{Z}$. Sorry, I don't know an elementary proof.

Improve this answer
$\endgroup$
1
  • $\begingroup$ That sounds pretty elementary to me, thanks. But diagonalizable means isomorphic to a subgroup of the diagonal matrices $\mathbf D_n$, right? So a subgroup of $\mathbb G^n_m$ is even a diagonal group, isn't it? Can you say a word about how to define this functor between the categories you mention? $\endgroup$
    – Drike
    Commented Feb 9, 2018 at 10:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .