I read the following in Fulton's book. Could you help me to solve it? Thanks!

Example: $V = V(XW - YZ) \in A^4(k)$. $\Gamma(V) = k[X, Y, Z, W]/(XW-YZ)$. Let $\overline{X}, \overline{Y}, \overline{Z}, \overline{W}$ be the residue 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 \neq 0 $ or $w \neq 0$.

Exercise 2-20: 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. Show that the pole set of $f$ is exactly $\{(x, y, z, w) \mid y = w =0\}$.


Suppose that $f=a/b$ with $a,b\in\Gamma(V)$. Note that $\Gamma(V)$ is an integral domain.

Then $a/b = \bar{X}/\bar{Y}$ thus in $k(V) = \mathrm{Frac}(\Gamma(V))$, we have $a\bar{Y} = b\bar{X}$, i.e. $aY-bX$ vanishes on $V$. Now, taking $P=(1,0,1,0)$, note that $P\in V$ and that $a(P)Y(P)=0$ but $X(P)\neq 0$. Hence $b(P)=0$.

  • $\begingroup$ Thanks you! I see your argument works for the second assertion. What's about the first one? $\endgroup$ Aug 16 '17 at 16:33
  • $\begingroup$ Here is how to prove the first part: suppose $b(x,y,z,w)\neq 0$ for all $y\neq 0$. It then follows that $b$ depends on $y$ alone (see this post for a simple argument). So if $b(x,y,z,w)\neq 0$ also for all $w\neq 0$, then $b$ is actually a constant. Since $zb-aw\in (xw-yz)$, evaluating at $(0,0,1,0)$ we find $b=0$, that is impossible. $\endgroup$
    – Math101
    Dec 17 '19 at 11:09
  • $\begingroup$ @Math101, Shouldn't it be evaluating at $(0,0,1,1)$? $\endgroup$
    – Steve
    Dec 10 '20 at 8:39

I don't think the above arguments for this problem are correct. A sketch solution is given by Mumford on page 21 of "The Red Book of Varieties and Schemes" (footnote 2).

Here are some details.

  1. $V = V( wx - yz )$ has $I(V) = (F)$ where $F = wx-yz$ is irreducible in $k[x,y,z,w]$. Let $U_1 = V - V(w), U_2 = V - V(y)$. Then $z/w$ is regular on $U_1$ and $x/y$ is regular on $U_2$. On $U_1 \cap U_2$ we have $z/w = x/y$ since $wx - yz \in I(V)$.

The most we can expect at this point is to have a rational function defined on $U_1 \cup U_2 = V - V(w,y)$.

  1. Let $Z = V(w,y) \subseteq V$. Let $a,b \in k[x,y,z,w]$ be such that $a/b$ is regular on $V - Z$. Note that $b(1,0,1,0) = 0$ is allowed, which explains why one of the solutions already given is incorrect. A correction suggests to consider $P=(0,0,1,1)$, but this does not lie on $V$.

Mumford argues that $V(b) \cap V = Z$ using a dimension argument. But the concept of dimension is not developed at this stage of Fulton's book. Instead, here is an elementary argument.

Since $a/b = x/y$ on $U_2$ we have $ay - bx = g(wx-yz)$ for some $g \in k[x,y,z,w]$. Re-arranging gives $y( a + gz ) = x(b + gw)$ and so $y | (b+gw)$ and so $b = -gw + hy$ for some $h \in k[x,y,z,w]$. It follows that $b$ is identically zero on $Z$.

  1. Now suppose $b$ is non-zero on $V - Z$. Define $Z' = V(x,z) \subseteq V$. Note that $Z \cap Z' = (0,0,0,0)$. Consider $b$ restricted to $Z'$. It is a polynomial $b(0,y,0,w) \in k[y,w]$ that is zero only at $(0,0)$. We can apply an argument along the lines of the one given above by Math101 (Dec 17, 2019) to get a contradiction (assuming $k$ is algebraically closed, or at least infinite). One could also argue using dimensions, since $V(b(0,y,0,w))$ should be co-dimension 1 in $k^2$.

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