Suppose $X$ is an integral scheme, and let $\eta \in X$ be its generic point. Then the local ring $\mathcal{O}_{X,\eta}$ is a field, called the function field of $X$ and denoted $K(X)$.

Why is $K(X)$ called a function field? In what sense (if any) are its elements functions? Is there some conceptual connection between regular functions (on a variety) and $K(X)$?

I'm asking because the name is quite suggestive, but it doesn't seem to follow directly from the definition that $K(X)$ is a field of "functions".


2 Answers 2


Yes, all elements of $K(X)$ are functions, but there is a caveat:

Caveat : the functions are not defined everywhere
The field $K(X)$ is simply the union of all $\mathcal O(U)$ with $U\subset X$ open, where one identifies $f\in \mathcal O(U)$ with $f|V\in \mathcal O(V)$ whenever $U\supset V$.
So I cheated a bit when I said that rational functions are functions: they are equivalence classes of functions.
But if you are an honest person, you can get along without cheating: in every equivalence class of such pairs $(U,f)$ there is one pair with $U$ maximal and by considering only such maximal pairs you may consider that $K(X)\subset \bigcup _{U \:\text {open}}\mathcal O(U)$ .
So, yes, $K(X)$ consists of regular functions, but these functions are not defined everywhere.
Shafarevich has made the interesting observation that this is a feature which distinguishes algebraic geometry from other geometric theories like topology or differential geometry.

  • $\begingroup$ OK, I reread the definition of the structure sheaf of an affine scheme from Hartshorne and it makes sense because the rings $\mathcal{O}(U)$ are defined explicitly as rings of functions. I think I was a bit confused because I have mainly been following Goertz&Wedhorn, who define the sheaf on principal open sets (as localizations of the coordinate ring) and then use sheaf properties to extend to arbitrary open sets. Thanks! $\endgroup$ Oct 16, 2013 at 20:50
  • $\begingroup$ Dear Balerion, you are welcome! $\endgroup$ Oct 16, 2013 at 21:10
  • $\begingroup$ Also, you are right that the elements of $\mathcal O(U)$ are functions, but with a strange codomain: I should have emphasized that aspect. $\endgroup$ Oct 16, 2013 at 21:22
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    $\begingroup$ Dear @jeffrey, yes it is true that elements of $\mathcal O_X(U)$ corespond to morphisms $U\to \mathbb A^1$. What I meant is that an element $f\in \mathcal O_X(U)$ can be seen as a function whose value at $x\in U$ is $f(x)\in \kappa(x)$, so that the codomain of $f$ is the union of all residue fields $\kappa(x)$. This is in contrast with classical algebraic geometry over $\mathbb C$, where functions on $U$ have all their values in the same field $\mathbb C$. This is because in a complex variety all closed points (the only ones considered classically) have residue field $\mathbb C$. $\endgroup$ Aug 2, 2014 at 8:22
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    $\begingroup$ Dear @jeffrey: A function $s\in \mathcal O_X(U)$ gives what you write but in my opinion this complicates things. Just refer to the definition of $s\in \mathcal O_X(U)$ in Hartshorne. It is given as a map $s:U\to \sqcup _{\mathfrak p\in U} A_\mathfrak p$ on page 70. This makes it visibly clear that $s$ has a strange codomain, viz a disjoint union of local rings. (After that you take residue fields in order to be closer to the classical point of view of seeing $\mathcal O_x(U)$ as functions) $\endgroup$ Aug 2, 2014 at 9:15

This might be kind of low-brow comparing to the other answer. Let us consider the simplest case first. Let $X=Spec A$, where $A=k[x]$. Then all the points in $X$ are given by the prime ideals of $A$, and since $A$ is a PID they are all generated by polynomials. Now the generic point corresponds to the minimal prime ideal of $A$, which is $\langle 0\rangle$ since $A$ is an integral domain. Therefore we have $$ \mathcal{O}_{X,0}\cong A_{0}=k(x) $$

Now we let $Y=Spec B$, where $B=k[x_{1}\cdots x_{n}]/I$ and $I$ is prime. Then the minimal prime ideal in $B$ is again given by $0$ as an equivalence class in the quotient ring. And similarly localization gives you the functional field on $Y$ to be the fractional field of $B$. If you think $Y$ as a complex algebraic variety, then this is the quotient field of the ring of regular functions.

So coming from an affine variety, to me the functional field can be thought as some kind of "fractional field", and the notation $K(X)$ is justified in this regard. I learned this perspective by reading Mumford's book. If you found any mistake, please point out to let me know.


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