Function on an algebraic curve: can a pole occur at a point in the projective closure of its zero set? Consider $x$ as a function on the affine $xy$-plane. Its zero set is a vertical line (the y-axis).  When this zero set is considered inside the projective plane, a single extra point needs to be added to obtain the projective closure, namely the point $P_\infty = (0:1:0)$.
The projective curve $y - x^2$ also contains the point $P_\infty$.  The function $x$ induces a function on this curve. But this induced function has a pole (of order 1) at $P_\infty$.
I am having trouble reconciling the fact that the induced function has a pole at a point where formerly (when considered as a function on the projective plane) it "almost" has a zero (i.e. at a point in the projective closure of the zero set).  Can anybody help to clarify the situation?
 A: You are looking at two different curves in the projective plane: first the curve $C_1 : x= 0$ (all points $[0:y:z]$) and then the curve $C_2 : yz = x^2$ (all points $[st:s^2:t^2]$).
The function you're calling $x$ on $\mathbf P^2$ is really $x/z$, which is not defined everywhere: it is only defined on the points $[x:y:z]$ where $z \not= 0$. So it is not defined at $P_\infty$. But if we only allow ourselves to approach $P_\infty$ along the curve $C_1$, then $x/z \to 0$ as points tend to $P_\infty$ on that curve. And if we only allow ourselves to approach $P_\infty$ along the curve $C_2$, then $x/z = st/t^2 = s/t \to \infty$ as points tend to $P_\infty$ on that curve (let $s \to 1$ and $t \to 0$).
Here is an analogous situation in $\mathbf R^2$. The function $y/x$ is not defined at the origin $O$, or more generally on the $y$-axis (where $x = 0$).  But if we focus on the line $L_m : y = mx$, which passes through $O$, then along this line the ratio $y/x$ has constant value $m$, so
$y/x \to m$ as $(x,y) \to O$ just on $L_m$. The function $y/x$ has different limiting values at $O$ when we approach that point along different lines, which is not a contradiction of anything.
Or think about $e^{-1/x^2}$ on $\mathbf R - \{0\}$. Although it is not initially defined at $x = 0$, it has a limiting value of $0$ as $x \to 0$ (from either direction).  And extending this to $x=0$ by setting its value there to be $0$ makes this an infinitely differentiable function at $x=0$ that is quite important: it leads to the notion of a bump function on the real line. Now extend the domain of this function to $\mathbf C - \{0\}$ by using the function of a complex variable $e^{-1/z^2}$ when $z \not= 0$. If $z  = x \to 0$ on the real axis, then $e^{-1/z^2} = e^{-1/x^2} \to 0$. But if $z = iy \to 0$ on the imaginary axis, then $e^{-1/z^2} = e^{1/y^2} \to \infty$.  So even though $e^{-1/x^2}$ is a smooth function at $0$ in real analysis, it makes no sense as function at $0$ in complex analysis (and it has an essential singularity there).
A: I think that the origin of this problem is that the function $f(x,y)=x$, that in terms of homogeneous coordinates really is $f(X,Y,Z)=X/Z$ is problematic at $P_\infty=(0:1:0)$. You see, neither $f$ nor its reciprocal is regular at $P_\infty$, and we cannot really discuss whether it has a pole or a zero at $P_\infty$, when we view $P_\infty$ as a point of $\Bbb{P}^2$.
On the other hand:

*

*when we view $f$ as a function on the zero locus of $X=0$, it becomes
the constant function zero, but

*when we view $f$ as a function on the parabola, in terms of the homogeneous coordinates as the zero locus $C$ of $YZ-X^2$, we have, from solving $YZ=X^2$, that
$$f=\frac X Z = \frac Y X,$$ when the latter form makes it plain that $f$ has a pole at $P_\infty$, when we view it in the function field of $C$.


At a smooth point $P$ of a curve $C$, the local ring $\mathcal{O}_P$ is a DVR, implying that every rational function $f$ in the function field $K(C)$ either
has a non-zero value at $P$, or a zero or a pole (of some order) there. All according to whether $f$ or $1/f$ or both belong to $\mathcal{O}_P$. On a higher dimensional variety we no longer have DVRs, and we run into this type of difficulties, when neither $f$ nor $1/f$ belongs to $\mathcal{O}_P$.

If we rewind back to multivariable calculus, it is easy to think of similar rational functions in several variables that have no limit at a given point. They may tend to zero, if you approach the point along some curve, but also tend to infinity, when we approach the point along some other curve.
A: Move from the affine plane in $\Bbb P^2$ defined by $z\neq0$ to the affine plane in $\Bbb P^2$ defined by $y\neq0$ (which basically amounts to adding $z$ in necessary places to homogenize everything first, then setting $y=1$ to inhomogenize afterwards). Now you can go back to using your intuition about $\Bbb R^2$ to see what's going on, as your $P_\infty$ has just become the origin of the $xz$-plane.
