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

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    $\begingroup$ 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 $\mathbb{C}$... $\endgroup$ – Eric Wofsey Mar 24 '16 at 2:26
  • $\begingroup$ @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. $\endgroup$ – John M. Campbell Mar 24 '16 at 2:36
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Any two algebraically closed fields of the same characteristic and transcendence degree are isomorphic. Brief proof: choose transcendence bases and a bijection between them, and then use the fact that an isomorphism between two fields can always be extended to an isomorphism between algebraic closures of the fields. (Note that this very heavily depends on the axiom of choice, and so you shouldn't expect to always actually be able to write down isomorphisms.)

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).

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