Question 2.20 W. Fulton

Question

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!

Answer 1

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.

Answer 2

Here is another solution which:

• Avoids using topological arguments to complete the pole set (as the OP asked for, since the base field is not assumed to be $\mathbb{C}$).
• Doesn't use the Nullstellensatz, and clears up some confusion I have about the other answer's usage of $V(b)\subseteq V(Y,W)$, since the polynomial defining $b$ may also vanish outside of $V(XW-YZ)$ while still leaving $a/b$ defined on all of $V(XW-YZ)\setminus V(Y,W)$.

The assumption on $k$ is simply that it is algebraically closed. I will write $V$ for $V(XW-YZ)$.

Note that $V(Y,W)\subseteq V$ and $V(X,Z)\subseteq V$, so given $b\in\Gamma(V)$ there are well-defined polynomials $b(X,0,Z,0)$ and $b(0,Y,0,W)$ obtained by restriction. Plugging in $Z=0$ we may also consider $b(X,0,0,0)$.

Supposing $f=a/b$, we have $b\overline{X}=a\overline{Y}$ and $b\overline{Z}=a\overline{W}$ in $\Gamma(V)$. Hence if $\overline{Y}$ and $\overline{W}$ both vanish, but one of $\overline{X}$ and $\overline{Z}$ doesn't, then $b$ must vanish. It remains to show that $b(0,0,0,0)=0$. But the polynomial $b(X,0,0,0)$ was just shown to have a root at every non-zero value of $X$, and $k$ is infinite (problem 1.6), so it must be the zero polynomial. This shows that the pole set of $f$ is all of $V(Y,W)$.

Now we want to show that $f$ cannot be defined on all of $V\setminus V(Y,W)$ by a single expression $a/b$. We know that $b(0,0,0,0)=0$, so the polynomial $b(0,Y,0,W)$ is either identically zero or non-constant. Now a non-constant polynomial in $k[Y,W]$ has an infinite vanishing set (problem 1.14), so there must be some non-zero choice of $(Y,W)$ (in fact, infinitely many) with $b(0,Y,0,W)=0$. This gives our desired point outside of the pole set (but still in $V$) where $b$ vanishes.