Pullback of divisor under the map $z\mapsto z^p$

Question

I'm a little confused right but I think this question can easily be answered.
Let $X\subset \mathbb C^3$ be the (affine) surface defined by $z=x^ay^b$, where $(x,y,z)$ are the coordinates on $\mathbb C^3$ and in this surface the divisor $D=\{z=0\}=a\cdot\{x=0\}+b\cdot\{y=0\}$ Consider the map $\phi: \mathbb C^3 \to \mathbb C^3, (x,y,z)\mapsto (x,y,z^p)$. What is $\phi^*D$?
I think it should be either $pD$ or $\frac{1}{p}D$ according to the following: the pullback of the equation $z=x^ay^b$ is $z^p=x^ay^b$. Now, either it is $pD$ because we have the $p$th power of $z$ or $\frac{1}{p}D$ because we can rewrite this as $z=x^{\frac{a}{p}}y^{\frac{b}{p}}$. Which argument (if one) is correct and why?

Answer 1

Actually, it turns out that the answer to my question is very trivial:

Let us distinguish the coordinates on the two $\mathbb C^3$'s and write the map $\phi$ as $\phi(x,y,w)=(x,y,w^p)=(x,y,z)$. Then $D=\{z=0\}=\{w^p=0\}=p\cdot\{w=0\}=p\cdot\phi^*D$.