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I have a question about a result of Abyankar, Heinzer, and Eakin, and a similar result in Russell. One of the results in the first paper is that if $ Y $ is a variety such that $ \mathbb{A}^{1}_{k} \times Y \cong \mathbb{A}^{2}_{k} $ then $ Y \cong \mathbb{A}^{1}_{k} $. I found an argument that suggests that this is not the case and so I was wondering if anyone could help me find what is wrong.

Let $ k $ be an algebraically closed field of characteristic $ p>0 $. Let $ \mathbb{G}_{a} $ act on $ \mathbb{A}^{2}_{k} $ as follows: \begin{align*} x_{1} & \mapsto x_{1}+t^{p} \\ x_{2} & \mapsto x_{2}-t^{p^{2}}+t. \end{align*}

The first part of this argument is the following claim:

If $ g(X) $ is the polynomial below: \begin{equation} g(X) = x_{1}^{p}+x_{2}, \end{equation} and $ f_{1}(X),f_{2}(X) $ are the following polynomials: \begin{align} f_{1}(X) &= x_{1}-g(X)^{p}, \\ f_{2}(X) &= x_{2}+g(X)^{p^{2}}-g(X), \end{align} then $ \mathbb{A}^{2}_{k} \cong \mathbb{G}_{a} \times \operatorname{Spec}(k[f_{1}(X),f_{2}(X)]) $.

Here is the p``r''oof of this claim:

If $ A $ is the ring $ k[f_{1}(X),f_{2}(X),t] $ then let $ \Phi $ be the following ring homomorphism from $ A $ to $ k[x_{1},x_{2}] $: \begin{align*} \Phi(f_{1}(X)) &= f_{1}(X) \\ \Phi(f_{2}(X)) &= f_{2}(X) \\ \Phi(t) &= g(X). \end{align*} If $ \Psi: k[x_{1},x_{2}] \to A $ is the following ring homomorphism: \begin{align*} \Psi(x_{1}) &= f_{1}(X)+t^{p} \\ \Psi(x_{2}) &= f_{2}(X)-t^{p^{2}}+t, \end{align*} then the following identities show that $ \Psi $ is an inverse of $ \Phi $: \begin{align*} \Phi(\Psi(x_{1})) &= \Phi(f_{1}(X)+t^{p}) \\ &= f_{1}(X)+g(X)^{p} \\ &= x_{1}-g(X)^{p}+g(X)^{p} \\ &= x_{1} \\ \Phi(\Psi(x_{2})) &= \Phi(f_{2}(X)-t^{p^{2}}+t) \\ &= f_{2}(X)-g(X)^{p^{2}}+g(X) \\ &= x_{2}+g(X)^{p^{2}}-g(X)-g(X)^{p^{2}}+g(X) \\ &= x_{2} \\ \Psi(\Phi(f_{1}(X))) &= \Psi(f_{1}(X)) \\ &= f_{1}(X) \\ \Psi(\Phi(f_{2}(X))) &= \Psi(f_{2}(X)) \\ &= f_{2}(X) \\ \Psi(\Phi(t)) &= \Psi(g(X)) \\ &= \Psi(x_{1}^{p}+x_{2}) \\ &= f_{1}(X)^{p}+t^{p^{2}}+f_{2}(X)-t^{p^{2}}+t \\ &= f_{1}(X)^{p}+f_{2}(X)+t \\ &= x_{1}^{p}-g(X)^{p^{2}}+x_{2}+g(X)^{p^{2}}-g(X)+t \\ &= x_{1}^{p}+x_{2}-g(X)+t \\ &= g(X)-g(X)+t \\ &=t. \end{align*} Therefore $ k[x_{1},x_{2}] \cong k[f_{1}(X),f_{2}(X),t] $.

The second part of the argument is the following claim.

If $ g(X), f_{1}(X) $ and $ f_{2}(X) $ are the following polynomials: \begin{align*} g(X) &= x_{1}^{p}+x_{2}, \\ f_{1}(X) &= x_{1}-g(X)^{p}, \\ f_{2}(X) &= x_{2}+g(X)^{p^{2}}-g(X), \end{align*} then $ k[f_{1}(X),f_{2}(X)] \cong k[y_{1},y_{2}]/\langle y_{1}^{p}+y_{2} \rangle $.

Here is the p``r''oof of this claim:

The following identities show that $ f_{1}(X)^{p}+f_{2}(X) $ is equal to zero: \begin{align*} f_{1}(X)^{p}+f_{2}(X) &= x_{1}^{p}-g(X)^{p^{2}}+x_{2}+g(X)^{p^{2}}-g(X) \\ &= x_{1}^{p}+x_{2}-g(X) \\ &= g(X)-g(X) \\ &=0. \end{align*} If $ \Phi: k[y_{1},y_{2}] \to k[f_{1}(X),f_{2}(X)] $ by sending $ y_{i} $ to $ f_{i}(X) $, then the kernel of $ \Phi $ contains $ \langle y_{1}^{p}+y_{2} \rangle $. Since $ \mathbb{G}_{a} \times \operatorname{Spec}(k[f_{1}(X),f_{2}(X)]) \cong \mathbb{A}^{2}_{k} $, the ring $ k[f_{1}(X),f_{2}(X)] $ must have dimension at least one. This means that the height of $ \ker(\Phi) $ must be less than two. The ring $ k[y_{1},y_{2}] $ is a UFD, and so the kernel of $ \Phi $ is principal. Therefore, the kernel of $ \Phi $ is exactly $ \langle y_{1}^{p}+y_{2} \rangle $.

This now seems to show that $ \operatorname{Spec}(k[y_{1},y_{2}]/\langle y_{1}^{p}+y_{2} \rangle) \times \mathbb{G}_{a} \cong \mathbb{A}^{2}_{k} $, but the variety $ \operatorname{Spec}(k[y_{1},y_{2}]/\langle y_{1}^{p}+y_{2} \rangle) $ is not isomorphic to $ \mathbb{A}^{1}_{k} $ as its field of fractions is a purely inseparable extension of degree $ p $ of the field of fractions of $ \mathbb{A}^{1}_{k} $.

If anyone sees the mistake, I would appreciate it if they would let me know.

Abhyankar, Shreeram S.; Heinzer, William; Eakin, Paul, On the uniqueness of the coefficient ring in a polynomial ring, J. Algebra 23, 310-342 (1972). ZBL0255.13008.

Russell, Peter, On affine-ruled rational surfaces, Math. Ann. 255, 287-302 (1981). ZBL0438.14024.

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  • $\begingroup$ "p``r''oof"?‍${}$ $\endgroup$
    – LSpice
    Mar 20, 2019 at 20:02
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    $\begingroup$ The $k$-algebra $k[y_1,y_2]/\langle y_1^p + y_2 \rangle$ is isomorphic as a $k$-algebra to the polynomial ring $k[y]$ via the homomorphism that sends $y_1$ to $y$ and that sends $y_2$ to $-y^p$. $\endgroup$ Mar 20, 2019 at 21:10
  • $\begingroup$ Thank you Jason!! I really appreciate it. $\endgroup$
    – schemer
    Mar 20, 2019 at 22:28
  • $\begingroup$ I only have one more question about this. If we changed the action to: \begin{align*} x_{1} & \mapsto x_{1}+t^{p^{2}}-t^{p}+t \\ x_{2} & \mapsto x_{2}+t^{p}, \end{align*} then I believe the argument changes $ g(X) $ to $ x_{1}^{p}+x_{1}-x_{2}^{p^{2}} $, $ f_{1}(X) $ to $ x_{1}-g(X)^{p^{2}}+g(X)^{p}-g(X) $ and $ f_{2}(X) $ to $ x_{2}-g(X)^{p} $. In this case, I believe that the ring $ k[y_{1},y_{2}]/\langle y_{1}^{p}+y_{2} \rangle $ changes to $ k[y_{1},y_{2}]/\langle y_{1}^{p}+y_{1}-y_{2}^{p^{2}} \rangle $. Is there a similar isomorphism in this case? Thank you. $\endgroup$
    – schemer
    Mar 20, 2019 at 23:09

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