TITLE: Hartshorne Lemma I.6.5; Why is $\mathfrak{m}_R\cap B\neq 0$?
QUESTION [8 upvotes]: I've been going back through some theorems in Hartshorne's Algebraic Geometry, trying to really understand the details.
I'm looking at Lemma I.6.5, which states (for those who don't have the book):

Let $K$ be a finitely generated field extension of transcendence degree one over $k$ ($k$ algebraically closed), and let $x \in K$. Then the set of discrete valuation rings of $K/k$ not containing $x$ is finite.

I'm following Hartshorne's proof, and I'm at the point where $y\in K$ is transcendental over $k$, and $B$, defined to be the integral closure of $k[y]$ in $K$, is a Dedekind domain and finitely generated as a $k$-algebra. 
He then supposes that $k[y]$ is contained in a DVR $R$ of $K/k$, which forces $B\subseteq R$ since $R$ is integrally closed in $K$. The next step is where I'm stuck:

If $\mathfrak{m}_R$ is the maximal ideal of $R$ then $\mathfrak{n}:=B\cap\mathfrak{m}_R$ is a maximal ideal of $B$.

Part of this is obvious: that is that $\mathfrak{n}$ is a prime ideal of $B$. As $B$ is a Dedekind domain, it suffices to show that $\mathfrak{n}$ is nonzero. I feel like there must be an obvious reason why this is the case, but I'm completely drawing a blank. I'd greatly appreciate it if someone could help me out here.

REPLY [8 votes]: If the pullback of $m_R$ is $(0)$, then $K = Frac(B) = B_{(0)} \subset R \subset K$, forcing $R = K$, which is absurd.