I am trying to understand a certain scenario and to do so, I sat down and calculated an explicit example. While doing so, I was able to "prove" two statements that directly contradict each other, which is (hopefully) not possible. I would like to ask you to help me find my mistake.
Let $n>2$. Consider the polynomial ring $A=\mathbb{C}[x,y]$ and the element $a:=x+2$. Set
$B := \mathbb{C}[x,y,u,v,w]/(u^n-x,v^n-y,w^n-a)$
Classical Understanding: $B$ is the coordinate ring of the affine variety
$Y=Z(u^n-x,v^n-y,w^n-x-2)\subset \mathbb{A}_{\mathbb{C}}^5$
and composing this inclusion with the projection $\mathbb{A}_{\mathbb{C}}^5\to\mathbb{A}_{\mathbb{C}}^2$ to the first two coordinates, we obtain a map $\pi:Y\to\mathbb{A}_{\mathbb{C}}^2$. For any $\zeta\in\sqrt[n]{2}\subset\mathbb{C}$, the point $(0,0,0,0,\zeta)$ is contained in $Y$, so there are at least $n$ preimages of the origin $(0,0)\in\mathbb{A}_{\mathbb{C}}^2$ under $\pi$. In fact, one can easily check that there are exactly $n$ preimages, but we shall not concern ourselves with this.
Modern Understanding: The canonical map $f:A\to B$ induces a morphism of varieties
$\pi:Y=\mathrm{Spec}(B)\to\mathrm{Spec}(A)=\mathbb{A}_{\mathbb{C}}^2$.
which is given by pulling back prime ideals via $f$. Let $P$ be the maximal ideal of $A$ generated by $x$ and $y$, hence the one corresponding to the origin. Assume that $Q\subset B$ is a prime ideal such that $f^{-1}(Q)=P$, i.e. $Q\in\pi^{-1}(P)$. Let us denote the image of $a\in\mathbb{C}[x,y,u,v,w]$ in $B$ by $\overline{a}$.
Since $x\in P$ and $f(x)\in f(P)\subset Q$, we have $\overline{x},\overline{y}\in Q$. Since $\overline{u}^n=\overline{x}$ and $Q$ is prime (hence radical), we have $\overline{u},\overline{v}\in Q$ and we observe that Q':=(\overline{u},\overline{v}) already satisfies f^{-1}(Q')=P. Hence, by the Going-Up theorem for integral extensions (Eisenbud Proposition 9.2), we know that Q' is maximal, hence Q'=Q. Alternatively, one can easily check that all elements of $B$ are mapped to elements of $\mathbb{C}$ under p:B\twoheadrightarrow B/Q': This is clear for $\overline{u}$, $\overline{v}$, $\overline{x}$ and $\overline{y}$, and we have
$p(\overline{w})^n=p(\overline{a})=p(2)=2$
so $p(\overline{w})$ is some $n$-th root of $2$.
Either way, the point $P$ has exactly one preimage under $\pi$, namely Q=Q'.
So, it is my understanding that I describe the same scenario in different languages, so the origin must have either one or $n$ preimages, but both can not be the case.