I need to prove that there is a valued field whose valuation group is $Q^2$. I know that the valuation group of $k((t^{1/n}))$ is $\frac{Z}{n}$ then if we take the union of $k((t^{1/n}))$ we get ${k((t))}_{pui}$ which has $Q$ as a valuation group. so in order to get a valuation group ${Q}^2$ can I look at ${k((t,s))}_{pui}$?


1 Answer 1


Construct a valuation on $k[x,y]$ with value group $\Bbb{Z}^2$ with the lexical order.

This valuation extends naturally to $\bigcup_{n\ge 1} k[x^{1/n},y^{1/n}]$ (with value group $\Bbb{Q}^2$ with lexical order) and to its fraction field.

  • $\begingroup$ Thank you. But is my answer incorrect? $\endgroup$ Mar 13 at 18:07
  • $\begingroup$ Yes because it takes 1 line to define the valuation! Unclear what you mean with $k((t,s))$ or its Puiseux series. $\endgroup$
    – reuns
    Mar 13 at 18:08
  • $\begingroup$ okay thank you. $\endgroup$ Mar 13 at 18:10
  • $\begingroup$ The field you'd probably want to consider first is $k((t))((s))$ (Laurent series in $s$ whose coefficients are Laurent series in $t$), note that it is not the fraction field of $k[[t,s]]$. and then $\bigcup_{n\ge 1}k((t^{1/n}))((s^{1/n}))$ $\endgroup$
    – reuns
    Mar 13 at 18:11
  • $\begingroup$ yes that's what I wanted to say I guess. sorry $\endgroup$ Mar 13 at 18:15

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