Intuition behind algebraic function field

Question

I don't think the definition of variety that you choose is important here, but in case you want mine, here it is: an algebraic variety over $k$ is a $k$-space $(X,\mathcal{O}_X)$ such that for all $x\in X$ there is an open $U\subset X$ such that $x\in U$ and $(U, \mathcal{O}_X|_U)$ is isomorphic (in $k$-spaces) to $(Y,\mathcal{O}_Y)$ with $Y\subset \mathbb{A}^n$ closed for some $n$, and $\mathcal{O}_Y$ the sheaf of regular functions.

Now that we've clarified this, let $(X,\mathcal{O}_X)$ be an algebraic variety. I have this definition for the function field of $X$: $$K(X)=\{ (U,f):U\subset X \text{ open dense }, f\in \mathcal{O}_X(U)\}\backslash\sim$$ where $(U,f)\sim(V,g)$ if and only if $f=g$ on $U\cap V$.

I am told that this is the fraction field of $\mathcal{O}_{X,P}$. I don't understand why this is true. Can someone help me see this?

Edit This is how I define $\mathcal{O}_{X,P}$. Pick an arbitrary $P\in U$. We define the local ring of $X$ at $P$ as follows:

$$\mathcal{O}_{X, P}:=\left\{(U, f): U \subset X\right. \text{ is open, } P \in U, \text{ and }\left.f \in \mathcal{O}_X(U)\right\} / \sim,$$

Please beware that I am not familiar with scheme theory.

Answer 1

This claim is only true when $X$ is irreducible, so we assume that was just omitted from the problem statement.

First we reduce to the case $X$ affine. Since $X$ is irreducible, any nonempty open subset is dense and any two nonempty open subsets meet, so if $X'\subset X$ is a nonempty open subset, then $K(X)\cong K(X')$ by the map $(U,f)\mapsto (U\cap X',f|_{U\cap X'})$ for $U\subset X$ and $(U',f')\mapsto(U',f')$ for $U'\subset X'$. Similarly, for any open $X'$ also containing $P$, we have $\mathcal{O}_{X,P}\cong\mathcal{O}_{X',P}$ via the map $(U,f)\mapsto (U\cap X',f|_{U\cap X'})$ for $U\subset X$ containing $P$ and $(U',f')\mapsto(U',f')$ for $U'\subset X'$ also containing $P$.

Next we show that for $X$ affine, $K(X)=\operatorname{Frac} \mathcal{O}_X(X)$. There's a map $\mathcal{O}_X(X)\to K(X)$ by $f\mapsto(X,f)$ which induces an injective morphism $\operatorname{Frac} \mathcal{O}_X(X)\to K(X)$, and we need to show this is surjective. Let $(U,f)\in K(X)$, and let $Z = X\setminus U$, the closed complement of $U$ in $X$. Then $Z=V(I)$ for some nonzero ideal $I\subset\mathcal{O}_X(X)$, and picking a nonzero $g\in I$, we have $\emptyset\neq D(g)\subset U$ . So we may assume $f\in\mathcal{O}_X(D(g))=(\mathcal{O}_X(X))_g$, and therefore every element of $K(X)$ can be represented as $f/g$ for $f,g\in\mathcal{O}_X(X)$ and $g\neq 0$. But that's exactly $\operatorname{Frac} \mathcal{O}_X(X)$.

Now since every $f\in\mathcal{O}_X(X)$ defines an element of $\mathcal{O}_{X,P}$ via $f\mapsto (X,f)$ and every $(U,f)\in\mathcal{O}_{X,P}$ defines an element of $K(X)$ via $(U,f)\mapsto(U,f)$, we have a chain of injective maps $\mathcal{O}_X(X)\to\mathcal{O}_{X,P}\to K(X)=\operatorname{Frac} \mathcal{O}_X(X)$ so that the composite is equal to the natural inclusion of $\mathcal{O}_X(X)$ in its field of fractions. But this implies that $\operatorname{Frac}\mathcal{O}_{X,P} = \operatorname{Frac} \mathcal{O}_X(X) = K(X)$ too.

Answer 2

Here I'm assuming $X$ irreducible so your definitions make sense.

I don't know what kind of intuition you are hoping for, but if you know some commutative algebra and category theory, the idea is that both the function field $K(X)$ and the stalk $\mathcal O_{X,p}$ at $p \in X$ are colimits of certain diagrams of restrictions.

In fact, for $\mathcal O_{X,p}$ we can think in a diagram of all open neighborhoods of $p$ with inclusion maps $U \to V$ whenever $p\in U$ and $U\subseteq V$. When we apply the contravariant functor $\mathcal O_X$ to this diagram, what we get is a diagram with all arrows reversed, all of then being restrictions. As in the case above, $$res_{V,U} : \mathcal O_X(V) \to \mathcal O_X(U)$$ Now, the thing is that since the diagram of these open sets with inclusions is a poset, then this system is filtered, that is, the colimit of this diagram is just the disjoint union indexed by open sets and followed by a quotient $$\frac{\bigsqcup_{p\in U} \mathcal O_X(U)}{\sim}$$ where for $f \in \mathcal O_X(U_1)$ and $g\in \mathcal O_X(U_2)$ we say $f\sim g$ if their restrictions are equal in $U_1\cap U_2$ (here we need that $U_1\cap U_2 \neq \emptyset$, that is guaranteed for irreducible $X$). The $k$-algebra structure in this object is defined by restricting all functions to the intersection of their respective open sets, and then computing the operations in this intersection.

For $K(X)$ the colimit is taken instead in the diagram of every nonempty open set of $X$.

Observing that we can always take an affine open $Y$ inside $X$ and irreducibility gives $Y$ dense in $X$, what we get is that we can change the indexation of our diagram from open sets $U$ of $X$ to open sets $Y\cap U$ in $Y$, and now we have the same as an irreducible affine variety. In the end, all the restrictions in this setting should be injective, as all of then are localizations on a finite number of functions and $\mathcal O_X(Y)$ is a domain. Naturally, any non-zero function $f\in \mathcal O_X(Y)$ becomes invertible in $K(Y) = K(X)$ because they are already invertible in the principal open $$Y_f := \{p \in Y\ |\ f(p) \ne 0\}.$$ What we get is that $K(X)$ is the fraction field of $\mathcal O_X(Y)$ (not $\mathcal O_X(X)$, try this for $X = \mathbb P^n$ for example). Also, since $\mathcal O_{X,p}$ is just $\mathcal O_{Y,p}$ if we took $Y$ with $p\in Y$, then it's also true that $\mathcal O_{Y,p}$ is an intermediary localization between $\mathcal O_X(Y)$ and $K(Y)$, and so $K(Y)$ is also the fraction field of $\mathcal O_{Y,p}$.