# Different definitions of regular functions (Gathmann and Mumford)

I've started reading algebraic geometry by myself just a few days ago, so I apologize in advance if this question is stupid or non-sensical.

In Mumford's Red Book, he defines regular functions (although he doesn't use this notation, I'm assuming that he is defining the same thing; please correct me if I am wrong) as follows:

Definition 4 Let $$X \subset k^n$$ be an irreducible algebraic set, and let $$R$$ be its affine coordinate ring. Since $$X$$ is irreducible, $$I(X)$$ is prime and $$R$$ is an integral domain. Let $$K$$ be its field of fractions. Recall that $$R$$ has been identified with a ring of functions on $$X$$. For $$x \in X$$, let $$m_x = \{f \in R | f(x) = 0\}$$. This is a maximal ideal, the kernel of the homomorphism $$R \to k$$ given by $$f \to f(x)$$. Let $$\underline{o}_x = R_{m_x}$$. We have then $$\underline{o}_x = \{f/g | f,g \in R, g(x) \ne 0\}$$. Now, for $$U$$ open in $$X$$, let $$\underline{o}_X(U) = \bigcap_{x \in U} \underline{o}_x.$$

I am assuming that $$\underline{o}_X(U)$$ is the set of regular functions on $$U$$ even though Mumford does not specify it. If I understand correctly, the rings $$R_{m_x}$$ are viewed by their embeddings in the field of fraction $$K$$, and the intersection is happening inside the field of fractions. Therefore, every element $$\phi$$ of $$\underline{o}_X(U)$$ is an element of the field of fractions $$K$$, and the way we view them as functions on $$U$$, say at a point $$p$$, is by first selecting $$f, g \in R$$ such that $$\phi = f/g$$ as elements of $$K$$ and $$g(p) \ne 0$$ when viewed as a function, and then defining $$\phi(p) := f(p)/g(p) \in K$$.

However, I became quite unsure about this after looking up other definitions. For example, Gathmann defines the set of regular functions on an open set $$U$$ of an affine variety $$X$$ as follows (I'll try to use the same notation as above here):

Definition 3.1 Let $$X$$ be an affine variety, and let $$U$$ be an open subset of $$X$$. A regular function on $$U$$ is a map $$\phi: U \to k$$ with the following property. For every $$a \in U$$, there are polynomial functions $$f, g \in R$$, with $$f(x) \ne 0$$, and $$\phi(x) = \frac{f(x)}{g(x)}$$ for all $$x$$ in an open subset $$U_a$$ with $$a \in U_a \subset U$$.

In this definition, the function $$\phi$$ is no longer required to be given by an element of $$K$$; we only know that it is locally given by a quotient of two polynomials, and I am wondering if this quotient can be different as elements of the field of fractions $$K$$ at different points of $$X$$. I would like to know if these two definitions are saying the same thing. To be more specific, my question is:

Given such a regular function $$\phi$$ in the sense of the second definition, can we find an element $$\phi' \in \underline{o}_{X}(U) \subset K$$, such that $$\phi = \phi'$$ as functions on $$U$$?

Thanks for reading this long post.

These two definitions are indeed the same. Recall the sheaf of regular functions $$\mathcal{O}_X$$, as defined in Gathmann's notes, and the following standard theorem.

Lemma 3.19 in Gathmann's notes: Let $$X$$ be an affine variety, $$x \in X$$ associated to the maximal ideal $$\mathfrak{m} \subset A(X)$$. Then, $$\mathcal{O}_{X, x} \cong A_{\mathfrak{m}}$$.

Using this theorem, it will suffice to show the following.

Claim: Let $$U \subset X$$ be an open subset. Then $$\mathcal{O}_X(U) = \bigcap_{x \in U} \mathcal{O}_{X, x}$$.

Proof: First, let's show that the restriction map $$\varphi \mapsto [(\varphi, U)] \in \mathcal{O}_{X, x}$$ is injective, so that we can view all the rings above as subrings of $$K$$, the fraction field of $$A(X)$$. This will also show the $$\subseteq$$ direction.

Let $$\varphi$$ and $$\psi$$ map to the same element of $$\mathcal{O}_{X, x}$$. Then, as functions they agree in a neighborhood $$V \subset U$$ of $$x$$. Then, $$\varphi - \psi$$ is a regular function on $$U$$ which vanishes on a dense open subset $$V$$ (since $$U$$ is irreducible in the subspace topology!) so it must vanish on all of $$U$$, since the vanishing locus of a regular function is closed.

(In Gathmann's notes, the above argument appears as the 'identity theorem for regular functions' in remark $$3.5$$.)

Next, we must show the reverse direction. Let $$\varphi \in K$$ be in the intersection above. Then, $$\varphi$$ is regular in a neighborhood of each $$x \in U$$, since $$\varphi \in \mathcal{O}_{X, x}$$. This implies that $$\varphi$$ is regular on $$U$$ by the definition of a regular function, which concludes the proof.

• Thank you so much! Just checking if I understand correctly: the fact that $X$ is irreducible is used in the second paragraph of the proof, right? Jun 16 at 20:14
• @KyawShinThant No problem! That's correct; it's needed to prove the identity theorem, and so that the intersection and equality above them make sense (as subrings of $K$). Jun 16 at 20:23