I am having trouble understanding something in Corollary I.6.6 of Hartshorne.
Let $K$ be a function field of dimension one over $k$ (by which he means a finitely generated extension of transcendence degree 1, and I also believe he is taking $k$ to be algebraically closed), then any discrete valuation ring of $K/k$ is isomorphic to the local ring of a point on some nonsingular affine curve.
Now as a proof, he takes some $R \subseteq K/k$ a DVR, and then takes $y \in R \setminus k$. He then says to construct the variety as in the proof of Lemma 6.5. What I don't understand is why we have to make sure $y$ is in $R \setminus K$. Taking the appropriate $y$ here will allow us to always ensure that $y \in \mathfrak{m}_R,$ the maximal ideal of $R$, which is what I think he wants to do.
In the construction of Lemma 6.5, he takes an element $y$ in a DVR $R$ of $K/k$. He then considers the integral closure of $k[y]$ in $K/k$, which he denotes by $B$. This winds up being a Dedekind domain and also a finitely generated $k$-algebra. Thus $B$ is the affine coordinate ring of some affine variety $Y$, which because $B$ is a Dedekind domain, happens to be a non-singular affine curve.
He then shows that $\mathfrak{n} = B \cap \mathfrak{m}_R$, where $\mathfrak{m}_R$ is the maximal ideal of $R$, is maximal in $B$ and hence that $B =R$. Now as $\mathfrak{n}$ is maximal, it corresponds to a point $P$ on $Y$. Then shouldn't the localization $B_\mathcal{n}$ be isomorphic to the local ring at the corresponding point $P$? I don't see the need to take $y \in \mathfrak{m}_R$ to ensure that this happens. It seems the only reason he takes $y \in \mathfrak{n}$ is to show the actual statement of the theorem, the actual construction of $B_\mathfrak{n}$ doesn't seem to hinge on $y$ actually being in $\mathfrak{n}$. I am maybe (likely) not understanding some subtlety in the construction but it seems like just taking $y \in R$ should be enough.