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 $0$. 



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


