A question on the proof of Valuative criterion of Properness in Hartshorne. I read Hartshorne's Algebraic Geometry and was stucked in page 102, in the proof of "Valuative Criterion of Properness".
Let me summarize what I tried to understand as follows:

Suppose $Y$ be a scheme, and let $y_0 \in Y$ be a specialization of $y_1 \in Y$.
  Let $\mathcal{O}$ be the local ring of $y_0$ on $\overline{\{y_1\}}$ with induced reduced schemes tructure. Then, the quotient field of $\mathcal{O}$ be $k(y_1)$.

I was confused in this point, because it seems to be $k(y_0)$. I think this modification wouldn't do any harm for the proof, because we can still make a morphism $k(y_1) \subseteq k(z_1)$ via $k(y_0) \subseteq k(z_1)$.
 A: The quotient field of a domain $\mathcal O$ is the localization $(\mathcal O\setminus\{0\})^{-1}\mathcal O$, while the residue field $k(y_0)$ is the quotient $\mathcal O/\mathfrak m$, where $\mathfrak m$ is the maximal ideal of the local ring $\mathcal O$.
Why is $\mathcal O$ a domain?
Note here the scheme structure on $\overline{\left\{y_1\right\}}$ is the reduced induced scheme structure, i.e. on an affine open neighborhood $U=\operatorname{Spec}(A)$ of $y_0$, which is also a neighborhood of $y_1$, if the closed subset $\overline{\left\{y_1\right\}}\cap U$ is defined by the (unique) radical ideal $I\triangleleft\mathcal O_Y(U)=A$, then the closed subscheme structure of $\overline{\left\{y_1\right\}}\cap U$ is the affine scheme $\operatorname{Spec}(A/I)$.
If $y_1\in U$ corresponds to the prime ideal $\mathfrak p\triangleleft A$, then the closed subset $\overline{\left\{y_1\right\}}\cap U$ is actually $V(\mathfrak p)$, thus $I=\mathfrak p$ and hence $A/\mathfrak p$ is a domain.
Then what is $\mathcal O$? Say $y_0\in \overline{\left\{y_1\right\}}\cap U$ corresponds to the prime ideal $\mathfrak q\triangleleft A/\mathfrak p$. Then $\mathcal O$ is actually the localization $(A/\mathfrak p)_{\mathfrak q}$. So $\mathcal O$ is a domain as a localization of a domain.
And it becomes obvious now that the quotient field of $\mathcal O$ is the quotient field of $A/\mathfrak p$, that is $$(A/\mathfrak p\setminus\left\{0\right\})^{-1}(A/\mathfrak p)=((A\setminus\mathfrak p)^{-1}A)\,/\,\mathfrak p(A\setminus\mathfrak p)^{-1}A.$$
(Here the equality uses that localization is an exact functor.)
And this is nothing but the residue field of $A$ at $\mathfrak p$, i.e. $k(y_1)$.

Hope this clears some confusion.
