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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?

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  • $\begingroup$ I think $Q_1$ and $Q_2$ are regular on $\mathbb{A}^2$, so the restriction to $A$ should be regular as well? $\endgroup$ – red_trumpet Mar 7 at 22:01
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$\newcommand{\spec}[1]{\mathrm{Spec}(#1)}$ $\newcommand{\Ohol}{\mathcal{O}}$ $\newcommand{\ideal}[1]{\mathfrak{#1}}$

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.

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