Over an algebraically closed field any automorphism of the affine line will extend uniquely to an automorphism of the projective line. Will that still be true if we work over a general (potentially non reduced) ring?
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4$\begingroup$ No, it's not. Take a field $K$ and put $R:=K[\varepsilon ]/(\varepsilon ^2)$. Then $x\mapsto x+\varepsilon x^2$ is an automorphism of $\mathbb{A}^1_R$ which does not extend to an automorphism of $\mathbb{P}^1_R$. $\endgroup$– abxDec 31, 2022 at 20:15
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3$\begingroup$ @abx Are you sure? It seems to me that when $\varepsilon^2=0$ then this is the fractional linear transformation $x\mapsto \frac{x}{1-\varepsilon x}$. (In the other affine line with coordinate $y=1/x$ it is $y\mapsto y-\varepsilon$.) $\endgroup$– Tom GoodwillieDec 31, 2022 at 21:54
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2$\begingroup$ @Tom Goodwillie: Oops, you are right of course! It should be $x\mapsto x+\varepsilon x^n$ with $n\geq 3$. $\endgroup$– abxJan 1, 2023 at 5:05
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$\begingroup$ Answers can be posted as answers :) $\endgroup$– Martin BrandenburgJan 1, 2023 at 12:30
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1$\begingroup$ Yes, this has been worked out (more generally for $\operatorname{Aut}(\mathbb{A}^n) $) by Shafarevitch, On some infinite-dimensional groups II, Math. USSR-Izvestyia, Vol. 18, 1982, pp. 185-194. $\endgroup$– abxJan 1, 2023 at 20:35
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