In this textbook on algebraic geometry, what is the difference between $ k[X]_{I(p)} $ and $ \mathcal{O}_{X, p} $? I am trying to understand what is the difference between $ k[X]_{I(p)} $ and $ \mathcal{O}_{X, p} $ in the textbook by Steven Dale's introduction to algebraic geometry:


In most textbooks on algebraic geometry, $\mathcal{O}_{X, p}$ is usually what we denote the localization of a affine variety (we use variety to mean irreducible here) at a point. But in this textbook, $
\mathcal{O}_{X, p}
$ is given another definition and the localization at a point is defined to be $ k[X]_{I(p)} $. But the notation suggests there are the same thing, and the definition also seems to suggest the same. Are these actually the same object as we have defined them here?
 A: "In most textbooks on algebraic geometry, $O_{X,p}$ is usually what we denote the localization of a affine variety (we use variety to mean irreducible here) at a point. But in this textbook, $O_{X,p}$ is given another definition and the localization at a point is defined to be $k[X]_{I(p)}$. But the notation suggests there are the same thing, and the definition also seems to suggest the same. Are these actually the same object as we have defined them here?"
Answer: There are maps
$$\phi: A(p):=k[X]_{I(p)} \rightarrow B(p):=\cup_{U\subseteq X,p \in U} \mathcal{O}_X(U)$$
(where we view $\mathcal{O}_X(U) \subseteq k(X)$ as a sub ring of the quotient field and take the union inside the set $k(X)$) and
$$\psi: B(p) \rightarrow A(p)$$
defined as follows:  An element $s \in A(p)$ is by definition an equivalence class of quotients $s=f/g$ where $g \notin I(p)$. Hence there is a (non empty) open set $p \in V(p)$ with $g(q)\neq 0$ for all $q\in V(p)$, and it follows $(V(p),f/g)\in \mathcal{O}_X(V(p))$ is a regular function. Hence we may define $\phi(s):=(V(p),f/g)$.  You must check this is a well defined map.
For the other direction: Let $t \in B(p)$, it follows $t \in \mathcal{O}_X(U)$ for some open set $p\in U$. Hence there is an open (non empty) subset $p\in V(p) \subseteq U$ where the restriction $s_{V(p)}$ satisfies $s_{V(p)}=f/g$
with $f,g$ polynomials and $g\neq 0$ on $V(p)$. This means $g \notin I(p)$ and we may define $\psi(U,s):=f/g \in A(p)$. You may check that these maps are well defined and inverses to each other, hence there is an isomorphism
$$k[X]_{I(p)} \cong \mathcal{O}_{X,p}.$$
