Let $K$ be a field. Let $\mathcal{O}_K$ be the intersection of all valuation rings with quotient field $K$. Can someone give an example of a field $K$ in which we don't have a bijection of sets: $$ \mathrm{Spec}(\mathcal{O}_K)\cong\{\text{valuation rings with quotient field }K\}\,, $$ or can someone provide a proof that we always have such a bijection of sets.
Examples
- If $K$ is a number field, then $\mathcal{O}_K$ is the integral closure of $\mathbb{Z}$ in $K$ and thus the ring of integers of $K$. The primes of the ring of integers correspond to (non-archimedian) valuations on $K$. We have a bijection as described given by $\mathfrak{p}\mapsto\mathcal{O}_{K,\,\mathfrak{p}}$ where $K$ corresponds to $\mathfrak{p}=(0)$.
- Let $A$ be a valuation ring with quotient field $K$. Then we have a bijection of sets $$ \mathrm{Spec}(A)\cong\{\text{valuation rings }B\text{ with quotient field }K\text{ such that }A\subseteq B\} $$ which is given by $\mathfrak{p}\mapsto A_\mathfrak{p}$. In particular if there exists a unique minimal valuation ring $A$ with quotient field $K$ (minimal among all valuation rings with quotient field $K$), then we have $\mathcal{O}_K=A$ and the desired bijection is satisfied.
The question is about if these two examples can be generalized to arbitrary fields $K$. If not I would be happy to see an example where the bijection fails to be true.
Any kind of help will be appreciated. Thank you in advance.