# 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.

$$\newcommand{\mm}{\mathfrak{m}}$$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.