Action of the $n$-th roots of unity on $mathbb{A}^2$

Question

The following appears as Example 1.9.5 (c) in Fulton's Intersection theory.

Let $X$ be the quotient of $mathbb{A}^2$ obtained by identifying $(s,0)$ with $(mu s,0)$ for all $n$-th roots of unity $mu$; equivalently, $$X = operatorname{Spec} K[s^n, st, s^2t, dotsc, s^{n-1}t, t].$$

I don't understand how the $n$-th roots of unity can only act on the line ${t = 0}$, what happens to a general point $(s, t)$? But if $$mu(s, t) = (mu s, t),$$ then I don't see why $scdot t$ should be an invariant.
Or is Fulton not talking about group actions here?

KReiser · Answer

It's not a group action on the whole of $Bbb A^2$, because $mu$ doesn't act on points with nonzero $t$-coordinate. It might help you to think about an example: taking $n=2$, we're looking at $operatorname{Spec} k[s^2,st,t]$. As $k[s^2,st,t]cong k[x,y,z]/(y^2-xz^2)$, we're looking at the Whitney umbrella, which you can think about as grabbing the left half-plane, twisting it, and shoving it through the right-half plane so the positive and negative $x$-axes are overlapping. We can see that when we do this, we preserve all the points which aren't on the $x$-axis.

Answered by KReiser on November 19, 2021

Action of the $n$-th roots of unity on $mathbb{A}^2$

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