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Let $k$ be a field (I do not mind to further assume that $k$ is of zero characteristic, if that will make things easier).

For $k[x_1,\ldots,x_n]$ it is known that the affine and triangular automorphisms generate the group of automorphisms of $k[x_1,\ldots,x_n]$, call it $G_n$, see, for example, van den Essen's book. It is also known that $G_2$ is a free amalgamated group, see, for example, Dick's paper.

My question: Is the group of $k$-automorphisms of $k[x,x^{-1}]$, call it $\hat{G_1}$, known? of $k[x,y,x^{-1},y^{-1}]$? or more generally, of $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$? (call it $\hat{G_n}$). By 'known' I mean if it is not difficult to find a set of generators similarly to the generators of $k[x_1,\ldots,x_n]$ (or perhaps, a free amalgamated structure of $\hat{G_2}$ similarly to the free amalgamated structure of $G_2$).

Thanks for any hints and comments.

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  • $\begingroup$ $(x,y)\mapsto (x+a,y)$ does not give an automorphism, since $x+a$ is not a unit. $\endgroup$ – Eric Wofsey Apr 13 '17 at 21:07
  • $\begingroup$ Oh, you are right! thanks for the correction. $\endgroup$ – user237522 Apr 13 '17 at 21:08
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    $\begingroup$ With $U,V \in k[x,y]$, $(x,y) \to (U,V)$ is an automorphism iff $x$ and $y \in k[U,V]$. But for $k[x,y,x^{-1},y^{-1}]$ it is different, we need $U,V$ to be units : with $U,V \in k[x,y,x^{-1},y^{-1}]^{\color{red}{\times}}$, $(x,y) \to (U,V)$ is an automorphism iff $x,y,x^{-1},y^{-1} \in k[U,V,U^{-1},V^{-1}]$ @EricWofsey $\endgroup$ – reuns Apr 13 '17 at 21:14
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These automorphism groups are much easier to understand than the automorphism groups of polynomial rings. Take $A=k[x,x^{-1}]$. If $\varphi:A\to A$ is an endomorphism, then $\varphi(x)$ is a unit of $A$. By degree considerations, it is easy to see that the only units of $A$ are monomials $ax^n$ for $a\in k^\times$ and $n\in\mathbb{Z}$. For $\varphi$ to be surjective, we must have $n=\pm 1$. Conversely, for any $a\in k^\times$ and $n=\pm 1$, $x\mapsto ax^n$ does give an automorphism. Explicitly, then, the group is a semidirect product of $k^\times$ ($x\mapsto ax$) and $\mathbb{Z}/2$ ($x\mapsto x^{-1}$), with $\mathbb{Z}/2$ acting on $k^\times$ by inversion.

With more variables the story is similar but more complicated. An endomorphism of $k[x_1,\dots,x_n,x_1^{-1},\dots,x_n^{-1}]$ must send each $x_i$ to a unit, and the only units are $ax_1^{d_1}\dots x_n^{d_n}$ for $a\in k^\times$, $d_1,\dots,d_n\in\mathbb{Z}$. Such an endomorphism is an automorphism iff the exponents $(d_1,\dots,d_n)$ for each of the $x_i$ form a basis of the abelian group $\mathbb{Z}^n$. It follows that the automorphism group is a semidirect product of $(k^\times)^n$ (automorphisms of the form $x_i\mapsto a_i x_i$) and $GL_n(\mathbb{Z}$) (automorphisms of the form $x_i\mapsto x_1^{d_{i1}}\dots x_n^{d_{in}}$ where $(d_{ij})\in GL_n(\mathbb{Z})$).

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  • $\begingroup$ Nice answer thanks! (Just to be sure, you have used the fact that for a Noetherian ring, a surjective endomorphism is injective?). $\endgroup$ – user237522 Apr 13 '17 at 21:18
  • $\begingroup$ Yes, that is one quick way to prove sufficiency of the condition to have an automorphism in the general case. $\endgroup$ – Eric Wofsey Apr 13 '17 at 21:19
  • $\begingroup$ A proof that a surjective endomorphism of a Noetherian ring is injective (hence an automorphism) can be found in Lemma 10.30.10 stacks.math.columbia.edu/tag/00FM $\endgroup$ – user237522 Apr 13 '17 at 22:44

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