# What is $\left(\overline{\mathbb{C}(t)}\right)^{\times}$?

In a previous question, I asked about the automorphism groups of the groups of units of various fields. In an answer to this question it is explained that $\mathbb{C}^{\times} \cong S^{1} \oplus \mathbb{R},$ where $S^{1}$ denotes the multiplicative group consisting of all complex numbers of absolute value $1$, since it is clear that the mapping from $S^{1} \oplus \mathbb{R}$ to the underlying multiplicative group of $\mathbb{C}$ whereby $(\theta, r) \mapsto \theta e^{r}$ is a group isomorphism. In this answer, it is also explained that $\mathbb{C}(t)^{\times} \cong S^{1} \oplus \mathbb{R} \oplus \bigoplus_{I} \mathbb{Z},$ where the expression $\bigoplus_{I} \mathbb{Z}$ denotes an uncountably infinite direct sum of copies of the additive group $\mathbb{Z}$ indexed by an index set $I$ of cardinality $\mathfrak{c}$.

Given the tower of fields $$\mathbb{C} \subseteq \mathbb{C}(t) \subseteq \overline{\mathbb{C}(t)},$$ where $\overline{\mathbb{C}(t)}$ denotes a fixed algebraic closure of $\mathbb{C}(t)$, it is thus natural to consider the group of units of $\overline{\mathbb{C}(t)}$, the field of algebraic functions of one variable over $\mathbb{C}$.

It is known that $\left(\overline{\mathbb{C}(t)}\right)^{\times}$ may be embedded in the field of Puiseux power series. There is a known construction of the algebraic closure of $\mathbb{C}(t)$ in terms of Nash functions described in Orderings in real rational functions field.

Using this construction, is it possible to evaluate the underlying multiplicative group of the algebraic closure of $\mathbb{C}(t)$, e.g., as a direct sum of 'well-known' groups? Also, it is natural to ask:

(1) What is $\left(\overline{\mathbb{C}(t_{1}, t_{2}, \ldots, t_{n})}\right)^{\times}$?

(2) What is $\left(\overline{\mathbb{C}(t_{1}, t_{2}, \ldots,)}\right)^{\times}$?

• Do you want some reasonably explicit description, or just a description of the isomorphism type using the axiom of choice? All algebraically closed fields of cardinality $\mathfrak{c}$ and characteristic $0$ are isomorphic to $\mathbb{C}$... – Eric Wofsey Mar 24 '16 at 2:26
• @EricWofsey Thanks for your comment. I was simply interested in a description of the isomorphism type (using the axiom of choice). I didn't realize that all algebraically closed fields of cardinality $\mathfrak{c}$ and characteristic $0$ are isomorphic to $\mathbb{C}$. You may want to consider adding a version of your comment as an answer. – John M. Campbell Mar 24 '16 at 2:36

Any field of rational functions over $\mathbb{C}$ in at most $\mathfrak{c}$ variables is a characteristic $0$ field of transcendence degree $\mathfrak{c}$, and so it follows that its algebraic closure is isomorphic to $\mathbb{C}$. So all the groups you are asking about are isomorphic to $\mathbb{C}^\times\cong S^1\oplus\mathbb{R}$ (though the isomorphisms are not at all canonical and may require the axiom of choice to construct).