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I'm in an algebraic geometry class and I've been asked, for homework, to prove that certain maps are regular/rational. I have no idea how to do this. We did prove that the only regular functions on $\mathbb{P}^n$ are constant functions, using the method of reducing regular functions to locally regular functions, but that was the extent of it. Here's an example:

Consider the Segre embedding $\varphi \colon \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ and let $Q$ be its image (i.e. $\{xw=yz\}$), let $p = (1:0:0:0)$, and let $H = \{x = 0\} = (\mathbb{A}_{(0)}^3)^c$. Define $\pi_p \colon Q \to H$ as the projection map from $p$ (i.e. given a point in $q \in Q$ join it with a line to $p$ and define $\pi_p(q)$ to be the point where that line intersects $H$). Prove that this map is rational. Then prove some other things (the other things are: $\pi_p$ maps the complement of the two lines $(a:0:b:0), (a:b:0:0)$ in $Q$ isomorphically to a copy of $\mathbb{A}^2$, and find the domain of $\pi_p$).

How do I even approach this problem? Given some $q = (a:b:c:d)$ with $ad=bc$ write down the line $\overline{pq} = (u + av : bv : cv : dv)$ and see where $u + av = 0$? This seems very "clunky" especially because our teacher just gave us a short lecture on how he would like to see "more optimal" solutions on the homework. Is there a better, more theoretical way to do this and, in general, how does one go about the problem of proving that some map is regular or rational?

The definitions I'm working with: Let $k$ algebraically closed. Let $X$ be an irreducible quasiprojective variety. A function $f \colon X \to k$ is regular at $x \in X$ if $f = \frac{u}{v}$ in a neighborhood of $x$, and $f$ is regular if it is regular at every such $x$. A map $f \colon X \to \mathbb{A}^n$ is regular if each of its coordinate functions is regular. If $Y$ is another quasiprojective variety then $f \colon X \to Y$ is regular iff $f \colon X \to \mathbb{A}^n_{(i)}$ is regular iff $f \colon f^{-1}(\mathbb{A}^n_{(i)}) \to \mathbb{A}^n_{(i)}$ is regular, where $\mathbb{A}^n_{(i)} = \{ (x_0:...:x_n) \mid x_i \neq 0\}$. Consider pairs $(U, f)$ where $U$ is open in $X$ and $f$ is a regular function, and define the equivalence relation $(U, f) \sim (V,g)$ if $g|_{U \cap V} \equiv f|_{U \cap V}$. Each such pair (or an equivalence class) is called a rational function and lives in $k(X)$, the function field of $X$.

Julien Clancy
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1 Answers1

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It's not clear to me that there will be a universal tip on how to go about these problems, but for future reference I'll post my answer to the specific example I mentioned in my question.

${\it Lemma.}$ If a function (resp. map) $f$ is homogeneous (resp. homogeneous of the same degree in each coordinate) then it is regular (or rational of the domain is not the whole space).

${\it Proof.}$ Suppose that $f = (f_0:...:f_n)$. Consider $A_i = f^{-1}(\mathbb{A}_{(i)}^n)$; we reduce to showing that $f \colon A \to \mathbb{A}_{(i)}^n$ is regular (or rational should we limit the domain). Here, the $i$th coordinate function $f_i$ is nonzero by definition of $A_i$. Thus we can rewrite $f$ as $f = (\frac{f_0}{f_i}:...\frac{f_n}{f_i})$. But this is obviously, in each coordinate, a quotient of homogeneous polynomials of the same degree. Thus is it regular/rational.

For the example I gave, it's easy to write down what that map does: $\pi_p(a:b:c:d)$ $= (0:b:c:d)$. So long as the point being projected is not $p$ this is fine. The set of points that are not $p$ forms an open set so we know by the lemma that $\pi_p$ acts rationally there.

Julien Clancy
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