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Let $\mathbb{G}_a$ be the additive group over an algebraically closed field $k$ of any characteristic. Let $X \to Y$ be a $\mathbb{G}_a$-torsor of $k$-schemes (of finite type - in case that is relevant). Suppose that $X$ is quasi-affine. Is it true that $Y$ is quasi-affine?

EDIT: As David pointed out below, one should also assume that $Y$ is separated.

ADD: Maybe I should add an explanation. The analogue assertion is not true for "affine" instead of "quasi-affine". The standard counterexample is $X = SL_2$ and $Y = SL_2/U$, where $U$ is the subgroup of unipotent upper triangular matrices ($Y$ is the $\mathbb{G}_m$-torsor corresponding to the tautological line bundle over $\mathbb{P}^1$). On the other hand, it is true that if $G$ is an affine group scheme of finite type and $U$ is a unipotent subgroup, that $G/U$ is always quasi-affine. In fact the quotient is quasi-affine if and only if $U$ is obervable in $G$, and $U$ is observable if it has no non-trivial characters (Referenz (not original): W. Ferrer Santos, A. Rittatore: Actions and Invariants of Algebraic Groups, Chapman & Hall (2005)).

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I'm not completely sure, but should this not follow from faithfully flat descent? –  Daniel Loughran Aug 13 '14 at 18:32
What kind of descent did you have in mind? –  Keerthi Madapusi Pera Aug 13 '14 at 18:48
Well as I said, I was not completely sure. It seems like Torsten's counter-example in the affine case shows however that descent probably won't work here. –  Daniel Loughran Aug 14 '14 at 7:01
By the way, Kemper has a result which lets us mostly ignore finite generation issues: There is a finitely generated subring $R$ of $\mathcal{O}(X)^G$ so that anything which is collapsed by the map $X \to \mathrm{Spec}(R)$ is also collapsed by $X \to \mathrm{Spec}(S)$ for any subring $S$ of $\mathcal{O}(X)^G$. So, although $\mathcal{O}(X)^G$ is not necessarily finitely generated, there are finitely generated subrings of it which are "good enough". See ams.org/mathscinet-getitem?mr=2532166 and the references therein, which will probably give a more accurate history than I have. –  David Speyer Aug 15 '14 at 16:14

2 Answers 2

up vote 3 down vote accepted

It's false! Take $\mathbb{A}^3$ with coordinates $(x,y,z)$. Blow up the origin, and let $E$ be the exceptional divisor. Delete the line $E \cap \{ z=0 \}$. The resulting quasi-projective variety will be our $Y$. Since $Y$ is quasi-projective, it is separated. $Y$ is an example of a quasi-projective variety with no complete curves that is not quasi-affine. I wrote an earlier answer showing that any $Y$ must have no complete curves and suggesting that might imply $Y$ quasi-affine; I have now deleted that answer since it was a dead end except to point me towards which $Y$ to investigate.

First, let us see that $Y$ is not quasi-affine. Since $B \ell_{(0,0,0)} \mathbb{A}^3$ is smooth, and $Y$ is obtained by deleting a codimension $2$ locus from it, any global function on $Y$ extends to $B \ell_{(0,0,0)} \mathbb{A}^3$. But any function on $B \ell_{(0,0,0)} \mathbb{A}^3$ contracts $E$. So any global function on $Y$ contracts $E \setminus \{ z=0 \}$, and global functions on $Y$ do not separate points.

Now, we build our torsor. We will cover $Y$ with two open charts: $U$ is the complement of the proper transform of $\{ z = 0 \}$. We have $U \cong \mathbb{A}^3$, with coordinates $(x z^{-1}, y z^{-1}, z)$; we define $p = x z^{-1}$ and $q = y z^{-1}$. The other chart will be $V:= Y \setminus E$. So $V \cong \mathbb{A}^3 \setminus \{ (0,0,0) \}$. The open set $V$ is not affine, but the global functions on $V$ are $k[x,y,z]$. Note that $U \cap V \cong \mathbb{A}^2 \times (\mathbb{A}^1 \setminus \{ 0 \})$, with coordinate ring $k[x,y,z^{\pm}]$.

Take $U \times \mathbb{A}^1$ and $V \times \mathbb{A}^1$, with coordinates $u$ and $v$ on the respective $\mathbb{A}^1$ factors, and let $\mathbb{G}_a$ act by translation on each $\mathbb{A}^1$. Glue these trivial torsors by $v = u + z^{-1}$; this makes sense since $z^{-1} \in \mathcal{O}(U \cap V)$. This will be $X$.

We write $\pi$ for the map $X \to Y$ and $\psi$ for the map $Y \to \mathbb{A}^3$, contracting $E$. To repeat, $X = \pi^{-1}(U) \cup \pi^{-1}(V)$. We have $\pi^{-1}(U) = \mathrm{Spec}\ k[p,q,z,u]$ and $\pi^{-1}(V) = \mathrm{Spec}\ k[x,y,z,v] \setminus \{ x=y=z=0 \}$

To show that $X$ is quasi-affine, consider the following $7$ functions: $$pz=x,\ qz=y,\ z,$$ $$a:=pzu+p=xv,\ b:= qzu+q = yv,\ c:=zu+1=zv,$$ $$d := zu^2+u = z v^2 - v .$$ For each function, I have given one formula which displays it as an element of $\mathcal{O}(\pi^{-1}(U))$ and another which displays it as an element of $\mathcal{O}(\pi^{-1}(V))$, so these are global functions on $Y$. Let $R$ be the ring generated by these functions. I claim that the map from $X$ into $\mathrm{Spec}(R)$ is an embedding (at which point it is an open embedding, since both varieties have dimension $4$).

We first check that the map separates points. Suppose that $\alpha_1$ and $\alpha_2$ are sent to the same point of $\mathrm{Spec}(R)$. Since $(x,y,z) \in R$, we see that $\psi(\pi(\alpha_1)) = \psi(\pi(\alpha_2))$. First, suppose this common point is not $(0,0,0)$. So $\alpha_1$ and $\alpha_2$ are in $\pi^{-1}(V)$ and are of the form $(x,y,z,v_1)$ and $(x,y,z,v_2)$. But we then have $x v_1 = x v_2$, $y v_1 = y v_2$ and $z v_1 = z v_2$, and $x$, $y$ and $z$ cannot all be $0$ on $V$, so $\alpha_1 = \alpha_2$ in this case. Now, assume that $\psi(\pi(\alpha_1)) = \psi(\pi(\alpha_2)) = (0,0,0)$. So $\alpha_1$ and $\alpha_2$ are in $U$, and are of the form $(p_1, q_1, 0, u_1)$ and $(p_2, q_2,0,u_2)$. But then $p_1 \cdot 0 \cdot u_1 + p_1 = p_2 \cdot 0 \cdot u_2 + p_2$, $q_1 \cdot 0 \cdot u_1 + q_1 = q_2 \cdot 0 \cdot u_2 + q_2$ and $0 \cdot u_1^2 + u_1 = 0 \cdot u_2^2 + u_2$, so we again conclude $\alpha_1 = \alpha_2$.

This shows that $X \to \mathrm{Spec}(R)$ is injective on points, but we want to know it is an embedding of schemes. Now that we know injectivity on points, we just need to do a local computation. The following table gives inverses on each of the open sets $\{ z \neq 0 \}$, $U \cap \{ zu+1 \neq 0 \}$, $V \cap \{ x \neq 0 \}$ and $V \cap \{ y \neq 0 \}$; we leave it to the reader to check that these sets form a cover. $$\begin{array}{@{\mbox{On}\ }r@{,\ }rcl} \{ z \neq 0 \} & (x,y,z,u) &=& (x,y,z, (c-1) z^{-1}) \\ U \cap \{ zu+1 \neq 0 \} & (p,q,z,u) &=& (ac^{-1}, b c^{-1}, z, d c^{-1}) \\ V \cap \{ x \neq 0 \} & (x,y,z,v) &=& (x,y,z,a x^{-1}) \\ V \cap \{ y \neq 0 \} & (x,y,z,v) &=& (x,y,z,b y^{-1}) \\ \end{array}$$

I suspect that a complete list of relations for $R$ is $$\mathrm{rank} \begin{pmatrix} a & b & c & d \\ x & y & z & c-1 \end{pmatrix} \leq 1$$ and I suspect that the image of $X$ in $\mathrm{Spec}(R)$ is everything except the line $\{ a=b=c=x=y=z=0 \}$. (On this line, $d$ is unconstrained.) Written this way, the $\mathbb{G}_a$ action is $$\begin{pmatrix} 1 & t \\ & 1 \end{pmatrix} \begin{pmatrix} a & b & c & d \\ x & y & z & c-1 \end{pmatrix} \begin{pmatrix} 1 & & & \\ & 1 & & \\ & & 1 & t \\ & & & 1 \end{pmatrix}.$$

But I doubt I will get around to checking this.

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Thank you very much! –  Torsten Wedhorn Aug 16 '14 at 5:45

No. Danielewski constructed examples where $X$ is a smooth hypersurface in $\mathbb{C}^3$ but the quotient $X/\mathbb{G}_a$ is the archetypical non-separated curve, a line with a double origin. (This was an auxiliary construction; his main goal was to construct surfaces $X$ and $X'$ such that $X \times \mathbb{G}_a \cong X' \times \mathbb{G}_a$ but $X \not \cong X'$. The idea is that, if $X$ and $X'$ are both $\mathbb{G}_a$ torsors over $Y$, then we automatically have $X \times \mathbb{G}_a \cong X \times_Y X' \cong X' \times \mathbb{G}_a$, but there is no particular reason we need to have $X \cong X'$ and Danielewski finds an example where we don't.) Danielewski's preprint was never published and doesn't seem to be online, but you can find good expositions by tom Dieck or Fieseler.

I will give Danielewski's simplest example. Set $X = \{ (x,y,z) : xz = y (y-1) \}$ and let $\mathbb{G}_a$ act by $(x,y,z) \mapsto (x, y+ax, z+a(2y-1)+a^2x)$. Note that any fixed point would have to satisfy $x=0$, $y=1/2$ and there is no such point on $X$, so this is a free action. I claim that the quotient is the line with doubled origin; the global functions on the quotient are $\mathbb{C}[x]$.

To understand why, it helps to describe $X$ in a different way. Take two copies of $\mathbb{C}^2$ with coordinates $(x_1, u)$ and $(x_2,v)$, and glue $\{ (x_1, u) : x_1 \neq 0 \}$ to $\{ (x_2, v) : x_2 \neq 0 \}$ by $(x_1,u) = (x_2, v+x_1^{-1})$. I claim that the result is isomorphic to $X$, with $x = x_1 = x_2$, $y = x_1 u = x_2 v +1$ and $z= x_1 u^2 - u = x_2 v^2 + v$. The above action restricts to the actions $(x_1, u) \mapsto (x_1, u+a)$ and $(x_2, v) \mapsto (x_2, v+a)$. So the quotient is two copies of $\mathbb{A}^1$, with coordinates $x_1$ and $x_2$, glued along everywhere but the origin. I leave the verification that $X$ really is the gluing of these two $\mathbb{A}^2$'s to you.

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Actually, I suspect that the answer might be yes, if you assume that $Y$ is separated. –  Alexander Braverman Aug 14 '14 at 19:06
@Alexander: In my situation I do know a priori that $Y$ is separated. Hence I would be very grateful if you could elaborate. –  Torsten Wedhorn Aug 14 '14 at 19:38
@David: Thank you very much for the example and the references. Of course, I will accept your answer. But I would like to wait a little hoping to attract more people to the modified question. –  Torsten Wedhorn Aug 14 '14 at 19:43
One strategy to prove that $Y$ is quasi-affine would be to show that $Y \to \mathrm{Spec}(A_0)$ is etale and of generic degree $1$, where $A_0$ is the ring of invariant functions. I think that should force an open immersion when $Y$ is separated, while having a good chance of being true even when $Y$ isn't. –  David Speyer Aug 14 '14 at 21:08

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