Construction of a Valuation Ring via Zorn's Lemma, except not

Question

In Atiyah & MacDonald they provide an abstract way to construct valuation rings, I'm curious how easy it is to work this out in general. As a refresher let $K$ be a field, $L$ an algebraically closed field and $\mathcal V$ be the set of ordered pairs $(A,f)$ where $A$ is a subring of $K$ and $f: A \rightarrow L$ is a homomorphism. In particular ordering $\mathcal V$ by $(A,f) \leq (A^\prime,f^\prime)$ if $A \subset A^\prime$ and $f^\prime |_A=f$, Atiyah & Macdonald show any maximal element of $\mathcal V$ is a valuation ring.

I was able to show that when $K=\mathbb Q$ and $L=\bar{\mathbb Z_p}$ that $\mathbb Z_{(p)}$ is a maximal element, are there any others? What are the maximal elements if we just take $K=\mathbb Q$ and $L=\bar{\mathbb Q}$. What if $K=\mathbb R$ and $L=\bar{\mathbb Z_2}$?

Answer 1

Note that the construction in A-M gives rise to a valuation ring whose maximal ideal is the kernel of $f: A\to L$. In particular, if $L$ has characteristic $p$, then the valuation of $p$ is positive or infinite.

If $K=\mathbb Q$, by Ostrowski's theorem, any valuation on $K$ is $\ell$-adic for some prime $\ell$. The above remark implies that the construction of A-M gives the p-adic valuation on $\mathbb Q$.

If $K=\mathbb Q$ and $L=\overline{\mathbb Q}$ or any field of characteristic $0$, there is a unique ring homomorphism from $K$ to $L$ and it is the injective. So the maximal element is just $A=\mathbb Q$ corresponding to the trivial valuaton.

When $K$ is any field of characteristic $0$ and $L$ has characteristic $p>0$, then the maximal elements are extensions of the $p$-adic valuation on $\mathbb Q$. As soon as $K$ has a transcendental element over $\mathbb Q$, there are infinitely many such valuations. Indeed, this is true over a purely transcendental subfield $\mathbb Q(t)$ of $K$ (valuation rings : $\mathbb Q[t]_{(t-r)\mathbb Q[t]}$, for any $r\in \mathbb Q$), and A-M's construction shows that any valuation of $\mathbb Q(t)\subseteq K$ extends to $K$.