Multiple Representations are Required to Identify the Pole Set of a Rational Function (2.4.20)

William Fulton Algebraic Curves, chapter 2, section 4, exercise 20:

Let $k$ be algebraically closed, and consider the rational function $f = x/y = z/w$ over the algebraic variety $V = \mathcal{Z}(xw-yz)$ in $\mathbb{A}^4(k)$. Show that the pole set of $f$ is precisely $\{(t_1,0,t_2,0)~|~t_1,t_2∈k\}$, and that it is impossible to write $f = a/b$ for $b(p)≠0$ for all $p∈V$ outside the pole set of $f$.

Proof: First of all, $y$ and $w$ are nonzero on points of $V$, so they are nonzero in $\Gamma(V)$. As well, $\mathcal{I}(\mathcal{Z}(xw-yz)) = (xw-yz)$ since $xw-yz$ is irreducible. Let $J_f = \{G ∈ \Gamma(V)~|~Gf ∈ \Gamma(V)\}$. It is clear that $J_f$ is an ideal in $\Gamma(V)$, and that $y,w∈ J_f$. Suppose $gf = u ∈ \Gamma(V)$ for some $g∈k[x,z]$; then we may write $$gx - uy = 0 \mod (xw-yz)$$ $$gx = 0 \mod (xw,y)$$ from which we conclude $g = 0$, so that $J_f = (y,w)$. This allows us to conclude that the denominator of $f$ in any way that we write it will always have $\{(t_1,0,t_2,0)\}$ as zeros, so that this is precisely the pole set of $f$. Suppose we could write $f = a / b$ with $b$ having zeros precisely at the pole set of $f$; this would be to say $$\mathcal{Z}(b,xw-yz) = \mathcal{Z}(y,w,xw-yz) = \mathcal(Z)(y,w)$$ $$\mathcal{Z}(b,xw-yz) ∩ \mathcal{Z}(x,z) = \mathcal{Z}(x,y,w,z)$$ $$\mathcal{Z}(b,x,z) = \{(0,0,0,0)\}$$ Since $b∈(y,w)⊆\Gamma(V)$, it will thus suffice to show that given $b ∈ (y,w) ⊆ k[y,w]$, we may find a zero besides $(0,0)$. Indeed, this is equivalent to showing $$\mathcal{Z}(b) = \mathcal{Z}(y,w)$$ is impossible. Indeed, every nonconstant polynomial in multiple variables over an algebraically closed field has infinitely many zeros (1.2.14), while $\mathcal{Z}(y,w)$ is just a single point.$~\square$