3
$\begingroup$

It is well-known that a definable field of finite Morley rank has no proper definable group of automorphisms (a proof can be found for example in the book "Stable groups" of Poizat). My question is: can something similar be said about morphisms of the additive group $K^+$ that are morphisms of a definable proper infinite subgroup of $K^{\times}$? Clearly we are in a higly pathological situation, indeed the field must be of characteristic $p$ (if not any additive morphism is a product by a scalar and therefore it must be the identity in our case) and a green field (since it must have a proper infinite definable multiplicative subgroup). Green fields of finite characteristic are unlikely to exist but I didn't find a proof of the non-existence of them.

New contributor
Moreno Invitti is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

1 Answer 1

1
$\begingroup$

Problem solved, they are exactly the automorphisms $\psi$ of $K$ such that $\psi(H)=H$. Indeed $K=\langle H\rangle_+$ (since $H^0$ is indecomposable) and so $\psi(xy)=\psi(x)\psi(y)$ for any $x,y\in K$.

New contributor
Moreno Invitti is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

You must log in to answer this question.

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