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?

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    $\begingroup$ 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


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


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