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Let $R$ be a ring and $R((x))$ the ring of formal Laurent series. The elements in the ring $R((x))$ are series of the form $$ f = \sum_{n\in\mathbb{Z}} a_n x^n, $$ where ${\displaystyle a_{n}=0}$ for all but finitely many negative indices $n$.

The spectrum of a ring is the set of all prime ideals of in the ring.

What is the spectrum of $R((x))$? Thank you very much.

Edit: Assume that $R$ is a field.

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    $\begingroup$ I’m confused; as you say, the spectrum of $R((x))$ is the collection of all prime ideals in $R((x))$. Are you looking for a description of these prime ideals in terms of the spectrum of $R$ or something along those lines? $\endgroup$
    – Alec Rhea
    Commented May 7, 2018 at 8:19
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    $\begingroup$ If you don't impose some restrictive properties on $R$, I doubt that an explicit description of the spectrum exists. $\endgroup$
    – Alex M.
    Commented May 7, 2018 at 9:08
  • $\begingroup$ @AlexM., thank you very much. I edited the post and assume that $R$ is a field. $\endgroup$
    – Jianrong Li
    Commented May 7, 2018 at 11:18
  • $\begingroup$ @AlecRhea, thank you very much. I edited the post and assume that $R$ is a field. $\endgroup$
    – Jianrong Li
    Commented May 7, 2018 at 11:18

1 Answer 1

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If $R$ is a field, then the same page that you link to states that $R((x))$ is a field, too - in fact, it is the field of fractions of the ring $R[[x]]$ of formal series. Since it is a field, its only ideals are $0$ and $R((x))$ - of which, by convention, only $0$ is considered prime. Therefore, the spectrum of $R((x))$ when $R$ is a field is reduced to a single point.

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