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Example - $V = V(XW-YZ) \subset \mathbb{A}^4(K)$. $\Gamma(V) = K[X,Y,Z,W]/(XW-YZ)$. Let $\overline{X}, \overline{Y}, \overline{Z}, \overline{W}$ be the residues of $X,Y,Z,W \in \Gamma(V)$. Then $\overline{X}/\overline{Y}=\overline{Z}/\overline{W} = f \in K(V)$ is defined at $P=(x,y,z,w)\in V$ if $y=0$ or $w=0$.

Problem - The question is to show that it is impossible to write $f=a/b$, where $a,b \in \Gamma(V)$, and $b(P)\neq 0$ for every $P$ where $f$ is defined. Furthermore must show that the set of poles of $f$ is exactly $\{(x,y,z,w)\in V | y=0\ \mbox{and} \ w=0\}$.

Observation: Cannot use topological arguments!

Although I have already seen this issue in Exercise 2-20 in Fulton's curves book it wasn't clear to me. Furthermore, it was not answered for the second part of the exercise. For the second part, I had the idea to do the following:

Taking $J_f = \{G \in K[X,Y,Z,W]|\ \overline{G}f \in \Gamma(V)\}$, show that $J_f = (Y,W)$ and then $V(J_f) = V(Y,W) = {y=w=0}$, i.e., $V(J_f) = \{(x,y,z,w)|\ y=0 \ \mbox{and}\ w=0 \}$, what is the pole set of $f$.

I would like ideas for the first part of any ideas to finish this second. I'm very grateful!

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I just copy and pasted this from my own personal notes on algebraic geometry.

$xw-yz=0$ then $\varphi=\frac{x}{y}=\frac{z}{w}$ in $k(V)$ and $f$ is defined on all points such that either $y$ or $z$ is not zero. Let us first prove that the domain of definition of $\varphi$ is $\{y\neq 0\}\cup \{w \neq 0\}$.

Assume that $\varphi=\frac{a}{b}$ then $$ay-bx=0$$ $$aw-bz=0$$ So we show that $y=0\wedge w=0\Rightarrow b=0$. As we have $$bx=0$$ $$bz=0$$ we see that on the $x,z$ plane $b=0$ (meaning $b$ with $y=0$ and $w=0$) except possibly at the origin. This implies that $b$ is zero on an open dense subset and so is zero everywhere. Thus we see that the domain of definiton is exactly $\{y\neq 0\}\cup \{w \neq 0\}$.

Now let us show that this rational function cannot be defined by a single rational expresssion. So again let us assume that $\varphi=\frac{a}{b}$ and further that $b=0\Rightarrow y=0\wedge w=0$ we derive a contradiction. The implication means that

$$V(b)\subseteq V(y,w)$$ which means that $$ I(V(y,w))\subseteq I(V(b))$$

or using the Nullstellensatz, $$\sqrt{(y,w)}\subseteq \sqrt{(b)}$$

since $y,w \in \sqrt{(y,w)}$ we have $y,w \in \sqrt{(b)}$ and so $b|y^m$ adn $b|w^m$ but this implies that $b$ is a constant and thus that $\varphi$ is everywhere defined, which is not the case.

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    $\begingroup$ I think you should replace all instances of $\cap$ with $\cup$ $\endgroup$ Jan 23 at 21:39
  • $\begingroup$ Thank you for your help. I think I understand your idea in general. In this first part, I can't use topology arguments, like density, if there's another way to do it would help a lot. In the second part, I liked the idea of ​​using the fact that the set of poles is $ (\{y=0\} \cup \{w=0\}) $ previously proven. However, I didn't understand why $b|y^m$ and $b|w^r$( I changed the exponent since they are not necessarily the same) implies b be constant. Could you help me with this? $\endgroup$ Jan 24 at 18:13
  • $\begingroup$ I have taken the liberty to replace your caps by cups, as quite correctly suggested by @Serguey Guminov. Mолодец, Сергей! $\endgroup$ Jan 27 at 16:22

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