# Easiest way to show these two functions are regular?

Set $$X = \mathbb{A}_K^2 = K^2$$, $$K$$ an algebraically closed field. I am considering $$A \, \colon= X - \{(0,0)\}$$ viewed as a locally closed subspace of $$X$$. By definition a locally closed subset is one that is the intersection of a Zariski open and a Zariski closed subset. Since $$A = V(\langle x,y \rangle)^c \cap X$$ the definition is satisfied, in particular $$A$$ is an open subspace. View $$A$$ as a quasi-affine algebraic variety, $$(A, \mathscr{O}_A)$$ with the structure induced from $$(X, \mathscr{O}_X)$$. If it matters, we know this space is irreducible since $$X$$ is irreducible and any open subspace of irreducible is also irreducible.

I would like to check that the functions $$Q_1 = x$$ and $$Q_2 = y$$ are regular on $$A$$. I feel like this should be extremely trivial, or obvious, but it is not. To be honest, the concept of regular functions have always been a bit slippery to me. My approach here would be to simply check that my definition of regular function is satisfied. My definition is $$Q_i$$ is regular on $$A$$, or $$Q_i \in \Gamma(A, \mathscr{O}_A)$$, if

1) $$Q_i \colon A \to k$$ is continuous,
2) if $$x \in A$$ and $$f \in \mathscr{O}_{x,K}$$ then $$f \circ Q_i \in \mathscr{O}_{x,A}.$$

Are there better ways, more intuitive ways, or easier ways to be thinking of regular functions so that the justification I seek is immediate?

• I think $Q_1$ and $Q_2$ are regular on $\mathbb{A}^2$, so the restriction to $A$ should be regular as well? – red_trumpet Mar 7 at 22:01

## 1 Answer

$$\newcommand{\spec}{\mathrm{Spec}(#1)}$$ $$\newcommand{\Ohol}{\mathcal{O}}$$ $$\newcommand{\ideal}{\mathfrak{#1}}$$

I must admit, that I do not understand the notation in point 2) of your definition of regular function, but I would propose you a definition that is both standard and easy to apply to your current example:

Let $$X = \spec{A}$$ be the prime spectrum of a commutative ring $$A$$ (with the Zariski-Topology of course). Then the first task is to define the sheaf of regular functions $$\Ohol_X$$ on $$X$$. A regular function $$s$$ on an open subset $$U \subseteq X$$ is then an element of $$\Ohol_X(U)$$.

So now to the definition: Take $$\Ohol_X(U)$$ to be the set of all functions $$s:U \to \coprod_{\ideal{p} \in U} A_\ideal{p}$$ with $$s(\ideal{p}) \in A_\ideal{p}$$. ($$\ideal{p}$$ is a prime of $$A$$, that is a point of $$X$$).

Furthermore require the following: For every $$\ideal{p} \in U$$, there is an $$f \in A$$ with $$f \notin \ideal{p}$$ and an element $$a/f^d \in A_f$$ such that for all $$\ideal{q}$$ with $$f \notin \ideal{q}$$ (that is $$\ideal{q} \in D(f)$$, to use a commonplace notation) we have $$s(\ideal{q}) = a/f^d$$ considered as element of $$A_\ideal{q}$$.

This sounds maybe like a bit scary pileup of notations, but it is actually a very natural definition. If $$A=k[x_1,\ldots,x_n]$$ is a polynomial ring this says that a regular function is given locally on $$D(f)$$ around $$P = \ideal{p}$$ by a quotient of polynomials $$g/h$$ , where the denominator $$h$$ does not vanish where $$f$$ does not vanish.

Now this definition above has two beautiful consequences: One can proof

$$(*) \quad \Ohol_X(D(f)) = A_f$$

for every $$f \in A$$ and

$$(**) \quad \Ohol_{X,\ideal{p}} = A_\ideal{p}$$

for every $$\ideal{p} \in A.$$

It should be now obvious that your function $$x, y \in k[x,y]=B$$ are regular in $$X=\spec{B}$$ and of course stay so in $$U = D(x) \cup D(y) \subseteq X$$.

You should really understand the proof of $$(*)$$ and $$(**)$$ (it is given in Hartshorne's "Algebraic Geometry") - in "practice" you will in most types of reasoning with regular functions use these isomorphisms.