Let $K$ be a complete discrete valuation field. That is, suppose that $K$ is a field, and that there is a map $\nu : K^{\times} \rightarrow \mathbb{Z}$ such that $\nu(a+ b) \geq \nu(a) \vee \nu(b)$, $\nu(1) = 0$, and $\nu(ab) = \nu(a) + \nu(b)$.
How might I show that there are no other valuations on this field?
For instance, when we complete $\mathbb{Q}$ to $\mathbb{Q}_p$, the other valuations $\nu_q$ disappear for $q \neq p$.
There are plenty of valuations on $\Bbb{Q}_p$ (for a cardinality reason $\overline{\Bbb{Q}}_p\cong \overline{\Bbb{Q}}_q$ when assuming the axiom of choice), but (up to a constant) there is only one discrete non-trivial valuation. This is because for $1+a p\in 1+p\Bbb{Z}_p$ then $(1+ap)^{1/n}\in \Bbb{Q}_p$ whenever $p\nmid n$, whence $v(1+ap)=0$ for $v$ to be discrete.
Then $$\Bbb{Q}_p^\times = \langle \zeta_{p-1}\rangle \times 1+p\Bbb{Z}_p \times p^\Bbb{Z}$$ Since $\zeta_{p-1}$ is a root of unity you'll have $v(\zeta_{p-1})=0$, for $v$ being non-trivial you need $v(p)\ne 0$, and $v(p)= v(1+\ldots+1)\ge v(1)=0$ gives $v(p)>0$ ie. up to a constant $v$ is the $p$-adic valuation.
It works the same way on any field complete for a discrete valuation $w$, the elements $a\in F$ with $w(a-1)>0$ still have $n$-th roots for $n$ coprime with the characteristic of the residue field so $v(a)=0$ for $v$ being discrete, then for $w(b)=0$ we must have $v(b)\ge 0$ since otherwise $v(1+\pi_F b^r)$ wouldn't be $0$ for $r$ large enough, and hence $v$ is determined by $v(\pi_F)$ which must be $\ge 0$ for the same reason.
