I can prove a Contradiction - Where's my mistake? 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.
 A: To begin with, note that the elements $u,v$ are not in $B$ so your definition of $Q'$ doesn't really make sense.    I suppose you mean $Q'=(\bar u,\bar v)\subset B$, and I'll assume that.   
The key to your paradox is quite simple: the ideal $Q'\subset B$ is not prime, and thus not maximal!
Indeed if we consider the projection $q:\mathbb C[x,y,u,v,w]\stackrel {def}{=}R \to B$ we get $q^{-1}(Q')=(x,y,u,v,w^n-2)\subset R$   which is clearly not prime.
Actually, if we call $Q(\zeta)\stackrel {def}{=}(\bar x,\bar y,\bar u,\bar v,\bar w-\zeta) \subset B $ , a maximal ideal,  we have $Q'=\bigcap \limits_{\zeta^n=2} Q(\zeta)=\prod \limits _{\zeta^n=2} Q(\zeta)$,  so that $V(Q')\subset Y$  consists of the $n$ reduced, rational points $Q(\zeta) \in Y$.
By the way, the schematic fiber $V(\bar x, \bar y)\subset Y$ of the origin of $\mathbb A^2_{\mathbb C}$ under $\pi $ is not reduced and has as reduction $V(Q')$.
Edit: Answer to question in comment. The surface $Y$ is indeed  smooth (it is isomorphic to the product of the affine $v$-line with the plane curve $w^n-u^n-2=0$ in the $w,u$-plane).
Hence  the ideal $Q(\zeta)$  can  be described by two generators near the point $Q(\zeta)$ , namely
 $\bar v$ and $\bar u$. If this seems strange remember that: 
$\bar x=\bar u^n,\bar y=\bar v^n $ and 
$\bar w-\zeta=\frac {\bar u^n}{\bar w^{n-1}+\ldots+ \zeta^{n-1}}$ in the local ring of the point  $Q(\zeta)\;$ 
[since $0=\bar w^n-\bar u^n-2=\bar w^n-\bar u^n- \zeta^n=(\bar w-\zeta)(\bar w^{n-1}+\ldots+ \zeta^{n-1})-\bar u^n \;$ ]
A: I don't think we know that the map $p$ is uniquely defined.
