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Let $K$ be a field and $K((x))$ be the formal Laurent series over $K$. The ring of formal power series $K[[x]]$ is a valuation ring of $K((x))$. I would like to see another concrete example of a valuation ring of $K((x))$ that is not comparable (by inclusion) with $K[[x]]$. Thanks for any help.

Mary
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  • Is it important to work in the field of formal Laurent series as opposed to a field of rational functions $K(x)$? – KCd Jan 26 '21 at 18:28
  • what is important to me is to see two incomparable valuation rings. The above question can be posed by replacing $K(x)$ by $K((x))$. – Mary Jan 26 '21 at 18:53
  • In that case consider $K[x]$ and $K[1/x]$. – KCd Jan 26 '21 at 20:22
  • Thanks for your answer. Is the intersection of these valuation rings, the field $K$? – Mary Jan 26 '21 at 21:27
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    Yes. But also I made an error: neither $K[x]$ nor $K[1/x]$ is a valuation ring. I should have used the local rings at 0 and at infinity: rational functions $f(x)/g(x)$ with $g(0) \not= 0$ and rational functions with $\deg f \leq \deg g$ (so a value at $\infty$ makes sense). Their intersection is quite big. For each absolute value $v$ on $K(x)$ that is trivial on $K$, there is a local ring $O_v$. The intersection of these over all $v$ is $K$. – KCd Jan 26 '21 at 21:44
  • Thanks for your time. So, I think this question on the field $K((x))$ to have more interesting examples. – Mary Jan 26 '21 at 22:00
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    Note that the field $K((x))$ has only one (up to equivalence) discrete valuation, compare https://math.stackexchange.com/a/4076344/96384 and discussion to https://mathoverflow.net/a/433926/27465. If, on the other hand, we allow general valuations, then every field except those contained in some $\mathbb F_p^{alg}$ have infinitely many inequivalent valuation rings, even for valuations of rank $1$. Compare https://math.stackexchange.com/q/4182677/96384. – Torsten Schoeneberg Nov 10 '22 at 17:52

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