Suppose you have two closed, irreducible subvarieties $Z\subseteq Y$ in some variety $X$. (I'm not sure if it matters, but for ease I'll just assume everything is over an algebraically closed field.) I know that there is some prime $\mathfrak{p}$ in the local ring $\mathcal{O}_{Z,X}$ corresponding to $Y$.
However, I don't understand why we have the isomorphisms $$(\mathcal{O}_{Z,X})_\mathfrak{p}\cong\mathcal{O}_{Y,X}\qquad\text{ and }\qquad\mathcal{O}_{Z,X}/\mathfrak{p}\cong\mathcal{O}_{Z,Y}. $$
Is there a nice proof of these particular isomorphisms? I'd appreciate reading them. Thank you.
We can and do reduce to the case where $X$ is affine , say $X=\operatorname {Spec}(A)$, and $Y\supset Z$ are integral subschemes corresponding to prime ideals $\mathfrak p\subset \mathfrak q$ of $A$.
Then $\mathcal O_{X,Y}=A_\mathfrak p$ and $\mathcal O_{X,Z}=A_\mathfrak q$.
Writing $\mathfrak p'=\mathfrak p A_\mathfrak q$, a prime ideal of $A_\mathfrak q$,the required equality $\mathcal O_{X,Y}=(\mathcal O_{X,Z})_{\mathfrak p'}$ boils down to the equality$A_\mathfrak p=(A_\mathfrak q)_{\mathfrak p'}$, which is §4, Corollary 4, page 24 in Matsumura's Commutative ring theory.
In the same vein we have $\mathcal O_Y(Y)=A/\mathfrak p$ and writing $\bar {\mathfrak q}=\mathfrak q\cdot(A/\mathfrak p)=\mathfrak q/\mathfrak p$ we obtain $(A/\mathfrak p)_{\bar {\mathfrak q}}=A_\mathfrak q/\mathfrak pA_\mathfrak q$ which translates as $(\mathcal O_Y(Y))_{\bar {\mathfrak q}}=\mathcal O_{X,Z}/\mathfrak p\mathcal O_{X,Z}$ or finally $\mathcal O_{Y,Z}=\mathcal O_{X,Z}/\mathfrak p\mathcal O_{X,Z}$