On page 130 of Hartshorne's Algebraic Geometry, a prime divisor on $X$ is defined as a closed integral subscheme $Y$ of codimension one. It is then claimed that the if $\eta \in Y$ is the generic point of $Y$, then the local ring $\mathcal{O}_{\eta, X}$ is a DVR with quotient field being equal to the function field of $X$. I was of the understanding that the local ring at the generic point of any integral scheme was always a field and that this field was \emph{defined} to be the function field. What am I missing here?
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1$\begingroup$ Seppi already answered your question. To complete the argument note that $O_{X,\eta}$ is a DVR since it's local, regular (noetherian) and has dimension 1. This follows from the equality $\operatorname{codim}(Y,X)=\dim O_{X,\eta}$. You can find a proof of the equivalence in 18.2 here: math.mit.edu/classes/18.782/LectureNotes18.pdf $\endgroup$– user347489Commented Jun 15, 2017 at 10:44
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Here Hartshorne is talking about the local ring of $\eta$ inside $X$, not $Y$. And $\eta$ is not the generic point of $X$.
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$\begingroup$ Would it be true if closed subscheme was replaced with open subscheme? I expect it would be trivially so then, right, since you could just take an affine open neighbourhood around the generic point inside that open subscheme and localize there. $\endgroup$– JoeCommented Jun 14, 2017 at 10:47
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1$\begingroup$ Yes, the generic point of a (nonempty) open subscheme of an integral scheme is the same as the generic point of the scheme. But that is a completely different setup: the original context was a closed subscheme of codimension 1. $\endgroup$– SeppiCommented Jun 14, 2017 at 10:50