symmetric monoidal (∞,1)-category of spectra
Let $K$ be a field. A Puiseux series with coefficients in $K$ is a formal Laurent series
where $r$ is a positive integer, $m$ is an integer, and each $a_k$ belongs to $K$. Somewhat more abstractly but also more meaningfully, if $\mathbb{N}_{\cdot}$ is the poset of positive integers ordered by divisibility (so that $1$ is the least element), then the field of Puiseux series is the filtered colimit of the diagram of fields
where each $F(n)$ is the field of Laurent series $K((t))$, but where $F(m) \to F(n)$ (in case $m|n$) is the field homomorphism taking $f(t)$ to $f(t^{n/m})$.
The field of Puiseux series carries a valuation whose values are in the ordered group of rationals $(\mathbb{Q}, +)$: for $f(t)$ as in (1), $v(f)$ is the least exponent $k/r$ for which $a_k \neq 0$.
Puiseux series were in essence considered by Isaac Newton, who developed a method of expanding algebraic functions as Puiseux series, based on an analogue of Newton's method of approximating roots. Here is a sample theorem:
If $K$ is algebraically closed and has characteristic 0, then the field of Puiseux series over $K$ is the algebraic closure of the field of Laurent series over $K$.
It is enough to show that every degree $n$ extension $E$ of the field of Laurent series $K((x))$ is of the form $K((x^{1/n}))$. For this, it suffices that the integral closure $B$ of $K[ [x] ]$ in $E$ be of the form $K[ [x^{1/n}] ]$.
Generally speaking, let $A$ be a complete DVR (discrete valuation ring) with maximal ideal $\mathfrak{m}_A$ and residue class field $k_A$, and let $F$ be its field of fractions. Let $E$ be a degree $n$ extension of $F$, and let $B$ be the integral closure of $A$ in $E$. Then $B$ is also a complete DVR. We may write the ideal $\mathfrak{m}_A B$ of $B$ as $\mathfrak{m}_B^e$ where $e$ is the ramification index, and we have
where the last equation holds because $\mathfrak{m}_B^i/\mathfrak{m}_B^{i+1} \cong k_B$ as $k_B$-modules and therefore also as $k_A$-modules. In the case $A = K[ [x] ]$ where $k_A = K$, we have that $dim_{k_A} k_B = 1$ since $K$ is algebraically closed, and therefore $e = n$. In other words, $(x)B = \mathfrak{m}_B^n$, so we can write $x = u \pi^n$ where $\pi$ generates the maximal ideal of $B$ and $u$ is a unit of $B$.
The residue class $\bar{u} \in K$ has an $n^{th}$ root (again by algebraic closure); in fact a simple root since $char(K) = 0$. By Hensel's lemma, this lifts to an $n^{th}$ root of $u$ in $B$. The element $u^{1/n} \pi$ is thus an $n^{th}$ root of $x$, and is a generator of the maximal ideal of $B$. Writing this element as $y = x^{1/n}$, the ring $K[ [y] ] = A[x^{1/n}] \hookrightarrow B$ is an $A$-submodule of full rank $n$ and integrally closed (being abstractly isomorphic to $A$, which is integrally closed because it’s a principal ideal domain and therefore a unique factorization domain), so that $K[ [y] ] = B$, as was to be shown.
(Intend to solve for $y$ in $y^3 - x y + 1 = 0$ as a Puiseux series in $x$.)
The sketched proof of theorem 1 was extracted from notes on a seminar by Boyarchenko on local class field theory:
and the reader may refer to the classic text by Serre for a fuller treatment:
For a noncommutative generalization see
Other references
We explore the concept of real tropical basis of an ideal in the field of real Puiseux series. We show explicit tropical bases of zero-dimensional real radical ideals, linear ideals and hypersurfaces coming from combinatorial patchworking. But we also show that there exist real radical ideals that do not admit a tropical basis. As an application, we show how to compute the set of singular points of a real tropical hypersurface. i.e. we compute the real tropical discriminant.
Last revised on July 2, 2016 at 11:39:24. See the history of this page for a list of all contributions to it.