Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)).

share|improve this question
    
I'm not completely sure, but should this not follow from faithfully flat descent? –  Daniel Loughran Aug 13 at 18:32
1  
What kind of descent did you have in mind? –  Keerthi Madapusi Pera Aug 13 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 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 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.

share|improve this answer
1  
Thank you very much! –  Torsten Wedhorn Aug 16 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.

share|improve this answer
    
Actually, I suspect that the answer might be yes, if you assume that $Y$ is separated. –  Alexander Braverman Aug 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 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 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 at 21:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.