I would like to know the number of valuation rings of $\Bbb Q_p((T))$.
I know $\Bbb Q_p$ has $2$ valuation rings, that is,$\Bbb Q_p$ and $\Bbb Z_p$.
Every algebraic extension of $\Bbb Q_p$ has more than $2$ valuation rings because of extension theorem on valuation.But $\Bbb Q_p((T))$ is not algebraic over $\Bbb Q_p$,I am at a loss.
How many valuation rings of $\Bbb Q_p((T))$ are there?
Thank you in advance.
Best Answer
There is only one (non-trivial) discrete valuation on $\Bbb{Q}_p((T))$.
For all $f\in 1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]]$ the binomial series gives that $f^{1/n}\in \Bbb{Q}_p((T))$ whenever $p\nmid n$. Therefore, that $v$ is discrete implies $v(f)=0$.
Decompose $$\Bbb{Q}_p((T))^\times = T^\Bbb{Z} p^\Bbb{Z} \langle \zeta_{p-1}\rangle \ (1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]])$$ Thus, it remains to find $v(\zeta_{p-1}),v(p)$ and $v(T)$.
Since $(\zeta_{p-1})^{p-1}=1$ we must have $v(\zeta_{p-1})=0$.
If $v(p)\ne 0$, then $p$ is a multiple of $1$ so we must have $v(p)>0$. For $r$ large enough we have $v(p^{-r} T)<0$ so that $v(1+p^{-r} T)<0$, a contradiction.
Whence $v(p)=0$. We must have $v(T)\ne 0$ for $v$ being non-trivial.
$v(T)<0$ would contradict $v(1+T)=0$ so we must have $v(T)>0$ and hence $$\{ x\in \Bbb{Q}_p((T))^\times,v(x)\ge 0\} = \Bbb{Q}_p[[T]]-0$$