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

Question

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

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.