A maximal ring wrt Domination is a valuation ring. Let $K$ be a field, and let $A$ be local subring of $K$, which is maximal wrt domination, ie for any local ring $B\subset K$, with $\mathfrak{m}_B\cap A=\mathfrak{m}_A$, we have $A=B. \ $ Then $A$ is a valuation ring.
My idea: assume there's $x \in K$ such that $x,x^{-1}$ are not in $A$. Consider $A[[x]]$, the ring of formal power series in $x$. It has unique max ideal $\mathfrak{m}_A+(x)$, so is local. Then taking $B=A[[x]]\cap K$ contradicts the maximality of $A$. I'm not $100\%$ sure that $B$ is a local ring, though.
 A: $\newcommand{\mm}{\mathfrak{m}}$Unless you have some notion of ring being "complete", it doesn't make sense to talk about the power series ring $A[\![x]\!]$ sitting inside $K$. (How do you make sense of $1 + x + x^2 + \cdots$?)

Here's how we can do it instead.
Claim. $A$ is normal (i.e., if $x \in K$ is integral over $A$, then $x \in A$).
Proof. Let $x \in K$ be integral over $A$. Then, $R = A[x]$ is an integral extension of $A$. Thus, there exists a maximal ideal $\mm_R \subset R$ that contracts to $\mm_A$. But then, $B = R_{\mm_R}$ is a local ring whose maximal ideal contracts to $\mm_A$. By maximality, we must have $A = R = B$. In particular, $x \in A$. $\Box$
Now we are ready to show that $A$ is a valuation ring. Let $x \in K \setminus A$. We wish to show $x^{-1} \in A$.
By hypothesis, the subring $R = A[x]$ is larger than $A$. This implies that $\mm_A R = R$. (Why?)
Thus, we can write
$$1 = a_0 + a_1 x + \cdots + a_n x^n$$
for some $a_i \in \mm_A$.
Now, note that $1 - a_0$ is a unit (in $A$) since it is outside $\mm_A$. Thus, we can rearrange the above to get
$$\frac{1}{x^n} = \frac{1}{1 - a_0}\left(\frac{a_1}{x^{n - 1}} + \cdots + a_n\right).$$
This shows that $x^{-1}$ is integral over $A$ and hence, $x^{-1} \in A$.

Here is how we can answer the "(Why?)": Note that we have the containment $\mm_A R \subset (1)$. This is containment as ideals of $R$.
Now, given any maximal ideal $\mm_R \subset R$, we see that $R_{\mm_R}$ is a strictly larger local ring than $A$. Thus, the maximal ideal of $R_{\mm_R}$ cannot contract to $\mm_A$. This means that $$\mm_A R_{\mm_R} = 1 R_{\mm_R}.$$
Thus, we have shown that the ideals $\mm_A R$ and $(1)$ agree at all localisations at maximal ideals. This is enough to show that they are equal to begin with.
