# Question 2.20 W. Fulton

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!

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.

• I think you should replace all instances of $\cap$ with $\cup$ Jan 23 at 21:39
• 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? Jan 24 at 18:13
• I have taken the liberty to replace your caps by cups, as quite correctly suggested by @Serguey Guminov. Mолодец, Сергей! Jan 27 at 16:22