# The existence of a valued field whose valuation group is ${Q}^2$

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 $${k((t,s))}_{pui}$$?

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.
• Yes because it takes 1 line to define the valuation! Unclear what you mean with $k((t,s))$ or its Puiseux series. Mar 13 at 18:08
• 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 $\bigcup_{n\ge 1}k((t^{1/n}))((s^{1/n}))$ Mar 13 at 18:11