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Algebraic Closure of Laurent Series

  1. Apr 21, 2006 #1

    AKG

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    Define:

    [tex]\mathbb{C}((t)) = \{t^{-n_0}\sum_{i=0}^{\infty}a_it^i\ :\ n_0 \in \mathbb{N}, a_i \in \mathbb{C}\}[/tex]

    What is its algebraic closure? My notes say that it is "close" to:

    [tex]\bigcup _{m \in \mathbb{N}}\mathbb{C}((t))(t^{1/m})[/tex]

    where [itex]\mathbb{C}((t))(t^{1/m})[/itex] is the extention of the field of Laurent series by the element t1/m. Is this in fact the closure? If not, what is it? Also, how would I prove that something is the algebraic closure of this field? I mean, if X is the algebraic closure, then one thing is to prove that every polynomial over C((t)) has a root in X, but how do I show that there is no intermediate field between X and C((t)) that also has this property? I.e. it's one thing to show that a field X contains the algebraic closure of another field Y, but how do I show that it IS the algebraic closure of Y?
     
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  3. Apr 21, 2006 #2

    Hurkyl

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    Can you write down a simple set of generators for your extension field, to simplify the problem?


    Or maybe use the ol' set theory trick: instead of trying to show that your extension field is an algebraic closure, try to show that it's a subset of an algebraic closure.
     
  4. Apr 21, 2006 #3

    AKG

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    For your first hint, maybe you mean {t1/p | p prime}, or even {t1/m | m natural}. And the algebraic closure would have to contain these, so the extention generated by these is contained in the algebraic closure. This actually wasn't that hard, and now that I think about it, I think something like this was in my notes. What's confusing me is that my professor said that the given thing was "close" to the algebraic closure, and he didn't feel certain enough to say for sure what the algebraic closure precisely is.

    So I think, given the above, it's relatively easy to show that the given extention is in the algebraic closure. But is the algebraic closure in it? That is, does every polynomial with coefficients in the set of Laurent series have a root in this union of one-element extentions?

    My professor sort of did this. He took an arbitrary polynomial. First he supposedly reduced the problem to one where the coefficients of the polynomial have no poles. A root x of this polynomial would be a Laurent-ish series in t1/m for some m. The coefficients of this series would be complex numbers. Using a method of undetermined coefficients approach, he gave an idea for how these coefficients could be determined. Is this the right idea?
     
  5. Apr 21, 2006 #4

    Hurkyl

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    Incidentally, these are called Puiseux series.

    Anyways, we're talking formal Laurent series, etc, right?

    It's easy to reduce the problem to where there are no "poles" through a change of variable -- make the substitution x = tn y for a sufficiently large n.


    It sounds like your professor's approach is just a variation on a basic theme: you want to incrementally construct a solution one term at a time (or more).


    My gut says that we want to prove the following conjecture:

    Let f(x, t) be an element of C[[t]] [x] of degree n in x. Then, the polynomial f(x, tn) has a solution in C[[t]] [x]. (Or maybe we want that exponent to be n!)


    And we want to find said solution by first constructing a solution modulo (t), and then modulo (t²), and so forth. Or, maybe we want to do it first modulo (tn), and then modulo (t2n), and so forth.

    Hensel's lemma might even solve the whole thing for us!
     
  6. Apr 22, 2006 #5

    mathwonk

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    this is a standard result in algebraic curves by robert walker, or is easy using complex analysis for convergent laurent series.
     
  7. Apr 25, 2006 #6

    mathwonk

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    thm 3.1, walker, algebraic curves, page 98, "the field of fractional power series is algebraically closed".
     
  8. Apr 25, 2006 #7

    mathwonk

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    the point here is essentially the existence of local parametrizations of any point of an algebraic plane curve, even a singular point.


    in the complex case, just project a nbhd of the point onto the z axis, and prove that in the complement of the singular point, it is a d sheeted covering space for some d.

    hence by classification of covering spaces of the punctured disc, this projection is equivalent to the holomorphic map z^d away from z=0.

    but then by the riemann extension theorem, this parametrization extends to the origin as well. i.e. the map z-->z^d lifts to a holomorphic map from the disc onto a nbhd of the singular point, which is an isomorphism from the punctured disc to a punctured nbhd of the singularity, and is also holomorphic at 0.


    the algebraic approach is to consider the equation of the plane curve as polynomial in w with coefficients which are polynomials in z, and then to ask for a "root" of this polynomial.

    such a root will be an expression w which is locally a dth root of a power series in z, or some such, i.e. lifting the d to 1 projection of the curve to a parametrization of the curve is like taking dth roots of the z values of the porojection.

    but this is always confusing to me.
     
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