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Let $k$ be a field and $(A,m)$ be the completion of the local ring of a smooth point of a $k$-variety. Let $x_1,x_2\in m\backslash m^2$ be regular elements. I am interested in knowing if one can find a $k$-linear automorphism of $A$ which takes the ideal $(x_1)$ to $(x_2)$.

If $k$ is perfect, it is easy to see that this can be done, since there is a $k$-linear embedding $A/m\hookrightarrow A$ which induces a $k$-linear isomorphism $$ A \cong (A/m)[|x_1,...,x_n|] $$ for any choice of regular sequence $(x_1,..,x_n)$ in $A$.

When $k$ is not perfect, I can hardly find any non-trivial $k$-automorphism of $(A,m)$.

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Here is a counterexample. Fix $a\in k\smallsetminus k^p$ and take the completion of the affine plane at the point $(0,a^{1/p})$. In other words, $A$ is the completion of the local ring $k[u,v]_{(u,v^p-a)}$.

Now take $x_1=u$ and $x_2=v^p-a$. Then $A/(x_1)$ is the completion of $k[v]$ at $(v^p-a)$, while $A/(x_2)\cong k(a^{1/p})[[u]]$. The residue field $k(a^{1/p})$ lifts to the second quotient, but not to the first, as you can check. Hence $A/(x_1)\not\cong A/(x_2)$ as $k$-algebras.

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  • $\begingroup$ I'm confused: it seems to me that $x_2=v$ does not belong to the prime ideal $(u,v^p-a)$ and thus is invertible in $A$. $\endgroup$ – YCor Apr 29 '15 at 14:35

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