# 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.

-
Minor quibble: $(0,0,\zeta,0,0)$ is not in $Y$, $(0,0,0,0,\zeta)$ is. – Craig Sep 8 '11 at 15:18
corrected, thanks. – Jesko Hüttenhain Sep 8 '11 at 15:27

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 $0=\bar w^n-\bar u^n-2=\bar w^n-\bar u^n- \zeta^n=(\bar w-\zeta)(\bar w^{n-1}+\ldots+ \zeta^{n-1})-\bar u^n \;$ ]

-
Your assumption is completely correct (I fixed that in the question), as is your answer! Thanks, that has relieved me from a major headache. I have another question, however: I think that $Y$ is regular, so $Q(\zeta)$ should be locally generated by two elements. Can you tell me which? – Jesko Hüttenhain Sep 8 '11 at 18:38
Dear @rattle, I have answered your question in an edit. You seem to have a very active way of learning algebraic geometry by linking the language of classical varieties to that of schemes and making interesting explicit calculations. Congratulations! – Georges Elencwajg Sep 8 '11 at 20:50
Ah, let $\sqrt[n]{2}=\{\zeta_1,\ldots,\zeta_n\}$. Then, locally around $Q(\zeta_i)$, the $\overline{w}-\zeta_j$ for $j\ne i$ become units, so $\prod_{j\ne i} \frac{1}{\overline{w}-\zeta_j} \cdot \overline{u}^n = \overline{w}-\zeta_i$ and therefore, the maximal ideal $\mathfrak{m}_{Q(\zeta_i)}$ is, in fact, generated by the images of $\overline{v}$ and $\overline{u}$. How ironic. – Jesko Hüttenhain Sep 8 '11 at 20:55
I swear, I figured it out mere seconds before you posted. Thanks a bunch, though, now I finally understand said scenario. – Jesko Hüttenhain Sep 8 '11 at 20:57
Dear@rattle: I completely believe you! It has often happened that the moment I post some text, I see that somebody else has just posted a practically identical text a few minutes before, which gives the impression that I have just copied it! – Georges Elencwajg Sep 8 '11 at 21:15

I don't think we know that the map $p$ is uniquely defined.

-
It's the projection to the quotient, I very much think it is. – Jesko Hüttenhain Sep 8 '11 at 15:30
You're right. Sorry. The problem is that the quotient is not $\mathbb{C}$, it's $\mathbb{C}[\hat{w}]/(\hat{w}^n -2)$, where $\hat{w}$ is the image of $w$ under the projection. – Craig Sep 8 '11 at 15:47
$\mathbb{C}[x]/(x^n-2)=\mathbb{C}$ since $\mathbb{C}$ is algebraically closed. – Jesko Hüttenhain Sep 8 '11 at 16:33
I don't see that that equality holds. The ideal $(x^n -2)$ consists of all polynomials in $x^n$, not all polynomials in $x$. So specifically, $\mathbb{C}[x]/(x^n-2)$ is an $n$-dimensional vector space over $\mathbb{C}$. – Craig Sep 8 '11 at 16:55
@rattle: $x^n - 2$ is not irreducible, so $(x^n - 2)$ is not maximal. – Qiaochu Yuan Sep 8 '11 at 17:39