If $f$ has a pole, does $f^2$ has a pole?

Question

I don't understand something in the exercise 2.17 of Algebraic Curves of Fulton.

Let $k = \overline{k}$ a field and $V$ be the variety defined by the zero of $ I = ( y^2 - x^2(x-1) ) \subset k[x,y]$.

Let $\overline{x}, \overline{y}$ be the coordinate functions. Then $z = \frac{\overline{y}}{\overline{x}}$ ìs a rational function with a pole at (0,0) but $z^2 = x-1$ and therefore has no poles on $\mathbb A^2_k$.

I don't understand how it's possible, because I tried to see poles exactly as in complex analysis (if $f$ has a pole at $z_0$ then $f^2$ too) but it seems not possible (or I made a mistake ...)

Answer 1

The main point is we are considering the restriction to $V$ of the function $z^2=\frac{\bar y^2}{\bar x ^2}$: that is the meaning when we put a bar over them. Inside the variety $V$, because of the defining equation $y^2-x^2(x-1)$, $z^2$ agrees with $x-1$ and is defined everywhere, thus has no poles.

Answer 2

The curve $V$ is not smooth at $(0,0)$. Around that point, your curve looks like $y^2=x^2$, which has two branches, one on which $y/x = +1$ and one on which $y/x = -1$. One way to think of what is happening is that the "function" $y/x$ has no limit as $(x,y) \to (0,0)$, but $(y/x)^2$ does. The behavior of $y/x$ around $(0,0)$ is not really the same kind of behavior as a "pole" in the sense of complex analysis.