# Algebraic closure of $\mathbb{C}(t)$

Let $$\mathbb{C}(t)$$ be the field of rational functions $$f(t) = \frac{p(t)}{q(t)}$$ with $$p,q\in\mathbb{C}[t]$$.

For instance, the function $$g(t) = \sqrt{t}$$ does not belong to $$\mathbb{C}(t)$$ but is lies in its algebraic closure $$\overline{\mathbb{C}(t)}$$ since it is a zero of the polynomial $$X^2-t\in\mathbb{C}(t)[X]$$.

Question. Is it correct to think of elements of $$\overline{\mathbb{C}(t)}$$ as functions $$h = h(t)$$ with values in $$\mathbb{C}$$ such that $$P(h(t)) = 0$$ for some $$P\in \mathbb{C}(t)[X]$$?

• Quite the definition, $P\ne 0$, the elements can be thought as functions analytic on $(0,r)$ for some $r>0$ depending on the function. The Galois group is known as the projective limit of homotopy classes of loops in $\Bbb{C}$, because the Galois conjugates of $h$ are its analytic continuations along closed-loops. – reuns Apr 4 at 20:28
• Almost by definition, an element of $\mathbb{C}(t)$ is a rational function on a curve $Y\to \mathbb{P}^1$. So, yes, you may think of it as multivalued function on the line satisfying a polynomial. – Donu Arapura Apr 4 at 21:38
• I meant "an element of $\overline{\mathbb{C}(t)}$ is..." – Donu Arapura Apr 4 at 23:58
• If an element of $\overline{C(t})$ is a function on a curve, then different elements are functions on different curves, and the question arises how do you add and multiply functions defined on different curves:-) – Alexandre Eremenko Apr 5 at 4:28
• I like @reuns 's idea of using the real arc germ $(0,r)$, it makes me think of Deligne's "tangential basepoints". In fact it is precisely that: it defines a point of the inverse limit of etale topoi of open subsets of $\mathbf{A}^1_\mathbf{C}$, which is the etale topos of $\operatorname{Spec} \mathbf{C}(t)$. – Piotr Achinger Apr 5 at 10:56

Elements of $$\overline{C(t)}$$ are not really functions since they do not have a common domain that would allow to add and multiply them. One way to think of $$\overline{C(t)}$$ is to consider the field of all formal Puiseux series centered at some point, so that addition and multiplication is well defined on them, and then take the subset consisting of those series which satisfy algebraic equations over $$C(t)$$. This definition depends on the starting point, but different points will lead to isomorphic fields which are different realizations of this algebraic closure.
• They do have a common domain of analyticity: $(0,\epsilon)$ – reuns Apr 5 at 9:06
• Also the Puiseux series of $f\in \overline{\Bbb{C}(t)}$ is for some $n$ a Laurent series in $t^{1/n}$ with finitely many negative coefficients and convergent for $|t|$ small enough. – reuns Apr 5 at 9:12
• @reins: 1. No, they dont! $\epsilon$ depends on the function. 2. Also notice that $t^{1/n}$ is not a function, even in a small neighborhood of $0$. So even as germs, algebraic functions have no common domain. – Alexandre Eremenko Apr 5 at 11:44
• It doesn't change the objection to your answer. A branch of $t^{1/n}$ is a function and when taking compatible branches $(t^{1/nm})^m = t^{1/n}$ the algebraic Puiseaux series are functions on $0<|t|<\epsilon$ with again $\epsilon$ depending on the element. – reuns Apr 5 at 11:52
• The objection is that for example you seem not to understand that any element of $\overline{\Bbb{C}(t)}$ is a function $\Bbb{R\to C\cup \infty}$, where at the poles and branch points we simply ask that the part on the right side is the analytic continuation of the part on the left side along a small half-circle on the upper half plane. In the obtained field of functions we extend by continuity whenever it is possible so that $\frac1{z}+ \frac{-1}{z}=0,z \frac1{z} = 1$. – reuns Apr 5 at 12:03