TITLE: Puiseux series over an algebraically closed field
QUESTION [10 upvotes]: Using the construction $R_n = K[t^\frac1n]$, $L_n = \text{Quot}(R_n)$ and $P = \bigcup_{n\in \mathbb{N}}L_N$ one automatically gets that the Puiseux series are a field. Nevertheless they are also an algebraically closed field if $K$ is an algebraically closed field. Why is that?

REPLY [7 votes]: As Pete Clark pointed out, you need $K$ to be of characteristic zero for this to be valid. To say that the Puiseux series are an algebraically closed field is to say that if $f(x,y) = \sum_{i=0}^n P_i(x)y^i$ is a polynomial whose coefficients are fractional Puiseux series $P_i(x) = \sum_{j = 0}^{\infty} a_{ij} x^{j \over N}$, with $P_n(x) = 1$, then one has a factorization
$$f(x,y) = \prod_{i=1}^n (y - Q_i(x))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$
Here the "roots" $Q_i(x)$ are also fractional Puiseux series over $K$, possibly with a larger $N$. 
The reason this is true is as follows. Newton's original method for finding Puiseux series for polynomials over ${\mathbb R}$ works over arbitrary $K$ of characteristic zero, in the sense that it produces a fractional Puiseux series $Q_1(x)$ such that $f(x,y)$ factors into $(y - Q_1(x))g(x,y)$ where now $g(x,y)$ is of the form $\sum_{i=0}^{n-1} R_i(x)y^i$ where the $R_i(x)$ are fractional Puiseux series. (If you think about it, there's a natural way to multiply $y - Q_1(x)$ and $g(x,y)$ together so this is well-defined.)
And it doesn't work just for polynomials, it works for all $f(x,y)$ of the above form $\sum_{i=0}^n P_i(x)y^i$ as well; $f(x,y)$ doesn't have to be a polynomial, and the presence of fractional powers don't play any major role since one is effectively just replacing the variable $x$ by the variable $x^{1 \over N}$ for some large $N$. 
Once you have $f(x,y) = (y - Q_1(x))g(x,y)$ you can then apply the procedure to $g(x,y)$ repeatedly until you have the factorization $(*)$.