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)).



Quasi-affineness of the base of a $\mathbb{G}_a$-torsor
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.
$$\begin{array}{@{\mbox{On}\ }r@{,\ }rcl}
\{ z \neq 0 \} & (x,y,z,u) &=& (x,y,z, (c-1) z^{-1}) \\
U \cap \{ zu+1 \neq 0 \} & (p,q,z,u) &=& (ac^{-1}, b c^{-1}, z, d c^{-1}) \\
V \cap \{ x \neq 0 \} & (x,y,z,v) &=& (x,y,z,a x^{-1}) \\
V \cap \{ y \neq 0 \} & (x,y,z,v) &=& (x,y,z,b y^{-1}) \\
\end{array}$$




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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. 


