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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.

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I didn't think about this carefully, but is $R$ an integral ring extension of $B$? If so, then you can use the going up theorem. If $\frak{m}_R$ contracts to the zero ideal, then take a maximal ideal of $B$, it contains $(0)$ and hence there is some prime ideal in $R$ containing $\frak{m}_R$ contracting to it contradicting maximality. –  Matt Jan 20 '13 at 22:18
    
@Matt, $R$ can't be an integral extension of $B$, can it? $B$, being the integral closure of $k[y]$ in $K$, is itself integrally closed in $K$. –  topspin1617 Jan 20 '13 at 22:20
    
Ah, yes. I didn't really read through the whole thing, but just look at the end for the question. –  Matt Jan 21 '13 at 1:33

1 Answer 1

up vote 7 down vote accepted

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

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Ah, yes; everything in $B\setminus \mathfrak{n}$ is already a unit in $R$, and so $B_\mathfrak{n}\subseteq R$. I had actually already made the observation that $\mathfrak{n}=(0)$ implies every element in $B$ becomes a unit in $R$ but completely missed the fact that that forces $K\subseteq R$. Thanks! –  topspin1617 Jan 20 '13 at 22:41

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