Let $X$ be an irreducible , affine variety, define the function field :
$$K(X) = \{(U,f)\mid U \subset X \text{ open }, f \in \mathcal{O}_X(U)\}/\sim$$
where $\mathcal O_X(U)$ denotes the ring of the set of regular functions defined on the open subset $U$.
Then we have the following injections:
$$\mathcal{O}_X(X) \to \mathcal{O}_{X,p}\to K(X)\tag{1}$$
Where $\mathcal{O}_{X,p}$ is the stalk at $p$, it can be shown that it's local ring (with the maximal ideal $m_{X,p} = \{(U,f)\mid f \in \mathcal{O}_X(U), f(p) = 0\}/\sim$), and $O_X(X)$ is identified with the coordinate ring, and it can be shown $K(X)$ is a field( the field of the fraction of $\mathcal{O}_X(X)$).
I want to prove that the localization of $\mathcal{O}_X(X)$ at maximal ideal $m_p\subset \mathcal{O}_X(X)$ at $p$ is identified with the stalk $\mathcal{O}_{X,p}$
I know by the universal property for the localization that there is a map $$\mathcal{O}_X(X) \to \mathcal{O}_{X}(X)_{m_p}\to K(X) \tag{2}$$
Localization at prime is also a local ring, However I don't know how to relate (1) to (2)?
Where regular function on $U$, are defined pointwise if exist some neighborhood $p\in V\subset U$ such that $f = \frac{g}{h}$ on $V$ for some non zero $h$.
and $\mathcal{O}_{X,p} = \{(U,f)\mid f \in \mathcal{O}_{X}(U)\}/ \sim$, with $(U,f) \sim (V,g)$ if exist some open $W\subset U\cap V$ such that $f|_W = g|_W$.
Best Answer
There are a lot of ways to see this identity, and each gives a different insight. In this answer, we'll give a very concrete way of proceeding (though, depending on your experience with category theory, how concrete it seems may vary).
We know
$$ \mathcal{O}_{p} = \big \{ (U,f) \mid f \in \mathcal{O}(U), p \in U \big \} \Big / (U,f) \sim (V,g) \text{ if } f \restriction_{U \cap V} = g \restriction_{U \cap V} $$
This is conveniently described as a direct limit. This is a worthwhile exercise, if you haven't seen it before.
$$ \mathcal{O}_p = \varinjlim_{p \in U} \mathcal{O}(U) $$
We would like to show this is equivalent to
$$ \mathcal{O}(X)_{m_p} = \mathcal{O}(X)[f^{-1} \mid f \not \in m_p] $$
By the universal property of localization, if we can show that each $f \not \in m_p$ is invertible in $\mathcal{O}_p$, then we'll get a (unique!) map $\mathcal{O}(X)_{m_p} \to \mathcal{O}_{p}$. If $f \mapsto (X,f)$ is the usual injection $\mathcal{O}(X) \to \mathcal{O}_p$, then our map will be given by
$$\frac{f}{g} \mapsto \left (U, \frac{f}{g} \right )$$
where $(U,g)$ is the inverse to $(X,g)$ we're assuming exists.
Conversely, using the universal property of a direct limit, if we can find maps $\varphi_U : \mathcal{O}(U) \to \mathcal{O}(X)_{m_p}$ which are compatible with the restriction maps $\mathcal{O}(U) \to \mathcal{O}(V)$, then we're guaranteed a (unique!) map $\mathcal{O}_p \to \mathcal{O}(X)_{m_p}$. Concretely, this will be given by
$$ (U,f) \mapsto \varphi_U(f) $$
which one can check is well defined on equivalence classes since the $\varphi_U$ are compatible with restriction maps.
Of course, both of these conditions will be satisfied. For instance, if $f \not \in m_p$ (that is, if "$f(p) \neq 0$") then $f$ is invertible in some neighborhood of $p$. Conversely, since every open set is a union of standard opens $D(f)$, we only need to build compatible maps $\mathcal{O}(D(f)) \to \mathcal{O}(X)_{m_p}$ whenever $p \in D(f)$. That is, we need maps $\mathcal{O}(X)_f \to \mathcal{O}(X)_{m_p}$ whenever $f \not \in m_p$. Here the usual inclusion works (and is clearly compatible with the restriction maps).
I'll leave it as an exercise that these maps are mutually inverse.
I hope this helps ^_^