6
$\begingroup$

I'm trying to understand how the various different types of series with coefficients in $\mathbb{C}$ relate to each other, and have read that the field of Hahn series $\mathbb{C}(x^\mathbb{Z})$ is basically the same as the Laurent series $\mathbb{C}((x))$. Hahn series are described as a generalisation of Puiseux series, and Puiseux series have rational exponents, so if one looks at the Hahn series in $\mathbb{C}(x^\mathbb{Q})$ with rational exponents, how does this relate to the complex Puiseux series?

According to Wikipedia, the field of (complex) Puiseux series in the indeterminate $x$ is defined to be the series that have complex coefficients and rational exponents with bounded denominator and the field $\mathbb{C}(x^\mathbb{Q})$ of Hahn series consists of series of the form $$\sum_{q\in\mathbb{Q}}c_qx^q,$$ where $\lbrace q\in\mathbb{Q}:c_q\neq 0\rbrace$ is well-ordered with respect to the natural ordering on $\mathbb{Q}$ and $c_q\in\mathbb{C}$ i.e. complex coefficients and exponents are contained in a well-ordered subset of $\mathbb{Q}$. (Please correct me if I have misinterpreted these definitions!)

They cannot be exactly the same since $$x^{1/2}+x^{2/3}+x^{3/4}+x^{4/5}+\cdots$$ is in $\mathbb{C}(x^\mathbb{Q})$ but is not a Puiseux series because the denominators are unbounded. Am I missing something?

$\endgroup$
1
  • 1
    $\begingroup$ A minor quibble: it's not that the terms $c_qx^q$ are well-ordered "with respect to the natural order on $\mathbb Q$", because that's not a "well-order". But this doesn't matter much. And, to reiterate the answer: no, you're not missing anything. I'd say a good part of this is just terminological, rather than conceptual. It is true that Puiseux series lie in algebraic extensions of the formal Laurent series ring $\mathbb C((x))$, while most Hahn series in $\mathbb C(x^{\mathbb Q})$ are not algebraic over $\mathbb C((x))$. $\endgroup$ Aug 10, 2021 at 19:11

1 Answer 1

4
$\begingroup$

A Hahn series is not a Puiseux series because the denominators in the exponents is generally unbounded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.