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Question: Let $X$ be a quasi-affine scheme of finite type over a finite field $k$ and let $A:=\Gamma(X,\mathcal{O}_X)$. Under which conditions is it true that $A$ is a finitely generated $k$-algebra?

As far as I understand, Amnon Neeman http://www.jstor.org/stable/2007052 gives in Remark 8.2 a counter example when $k=\mathbb{C}$ by using complex analytic techniques.

I am mainly interested in the case where $S=\mathbb{A}^1_k=\text{Spec }k[x]$, where $G$ is a closed subgroup scheme of the general linear group $GL_{n,S}$ over $S$, such that $G$ is flat over $S$ and such that $X:=GL_{n,S}\,/G$ is a quasi-affine scheme. Note that such an $X$ always is a scheme by Theorem 4.C of S. Anantharaman http://numdam.org/numdam-bin/fitem?id=MSMF_1973__33__5_0 and is smooth and of finite type over $S$ by [SGA3, VI$_{\rm B}$, Proposition 9.2(xii)].

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    $\begingroup$ A quasi-affine example with $A$ not of finite type is given by purely algebraic (rather than analytic) means in math.stanford.edu/~vakil/files/nonfg.pdf over any field $k$ such that there exists an elliptic curve $E$ over $k$ with $E(k)$ not a torsion group (so not for $k$ algebraic over a finite field). $\endgroup$ – grghxy Jun 16 '15 at 12:23
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    $\begingroup$ There are much more elementary examples than that one. Let $Y$ be the affine scheme $\text{Spec}(k[s,t,u]/\langle tu \rangle)$. Let $Z$ be the closed subset $\text{Zero}(s,t)$. Let $X$ be the quasi-affine complement $Y\setminus Z$. For every $n\in \mathbb{N}$, the element $s^{-n}u$ is an element in $\Gamma(X,\mathcal{O}_X)$. $\endgroup$ – Jason Starr Jun 16 '15 at 13:06
  • $\begingroup$ Dear Colleagues, thank you for your counterexamples. Are there any results on the positive side, guaranteeing that $\Gamma(X,\mathcal{O}_X)$ is a finitely generated $k$-algebra? $\endgroup$ – Urs Hartl Jun 23 '15 at 8:27
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Here are some additional details for the example above. First of all, it is homogeneous for a linear algebraic group scheme over $W = \text{Spec}(k[t,u]/\langle tu \rangle)$. Since $W$ is itself finite and flat over $S=\text{Spec}(k[v])$, it is tempting to apply Weil restriction to try and produce an example over $\mathbb{A}^1_k$. However, this does not work (at least not in the most obvious way). This illustrates an interesting feature of Weil restriction: it can sometimes transform quasi-affine schemes to affine schemes.

As above, let $Y$ be the affine scheme $\mathbb{A}^1_W = \text{Spec}(k[s,t,u]/\langle tu \rangle)$. Let $Z$ be the closed subscheme associated to the ideal $\langle s,t \rangle$. Let $X$ be the quasi-affine complement $Y\setminus Z$. By working with the covering of $X$ by the basic open affines $D(s)\subset Y$ and $D(t)\subset Y$, it is straightforward to compute $$ \Gamma(X,\mathcal{O}_X) = (k[s,t,u]/\langle tu \rangle)[r_1,r_2,r_3,\dots,r_n,\dots]/\langle tr_n, s^nr_n - u, s^mr_{m+n} - r_n \rangle, $$
where the restriction of $r_n$ to $D(s)$ is $s^{-n}u$ and where the restriction of $r_n$ to $D(t)$ is $0$. In particular, $\Gamma(X,\mathcal{O}_X)$ is not a finite type $k$-algebra.

To see that $X$ is homogeneous, let $H_W$ be the $W$-group scheme $H\times_{\text{Spec}(k)} T$, where $H$ is the closed $k$-subgroup scheme $\text{Spec}(k[a,a^{-1},b])$ of $\mathbf{GL}_{2,k}$ of matrices of the form $$\left( \begin{array}{rr} a & b \\ 0 & 1 \end{array} \right). $$ A $W$-action of $H_W$ on $Y$ is the same as a $k$-action of $H$ on $Y$ such that projection from $Y$ to $W$ is $H$-invariant. Define the action by $$ \left( \begin{array}{rr} a & b \\ 0 & 1 \end{array} \right)\cdot (s,t,u) = (as+bt,t,u). $$ This action preserves the closed subscheme $Z$. Thus it induces an action on $X$. There is a section of the projection $X\to W$, namely $(t,u) \mapsto (1,t,u)$. The stabilizer is the closed $W$-subgroup scheme $G_W\subset H_W$ that is the zero scheme of $a+bt-1$. This is $W$-flat. Thus, as a $W$-scheme, $X$ is the quotient $H_W/G_W$. The scheme $X$ is finite type, quasi-affine and smooth over $W$, but the associated affine scheme $\text{Spec}\Gamma(X,\mathcal{O}_X)$ is not finite type over $W$.

We could try to produce such an example over $S=\text{Spec}(k[v])$ from this example by taking the Weil restriction of $X$ with respect to the finite, flat morphism $$ \phi:W\to S, \ \phi^*v = t + u. $$ However, the Weil restriction of $X$ with respect to $\phi$ is actually affine. The Weil restriction of $\mathbb{A}^1_W$ with respect to $\phi$ is isomorphic to $\mathbb{A}^2_S = \text{Spec}(k[s_0,s_1,v])$, where the associated universal $W$-morphism $\psi:\mathbb{A}^2_S\times_{S,\phi} W \to \mathbb{A}^1_W$ is $$ \psi:\mathbb{A}^2_W \to \mathbb{A}^1_W, \ \ \psi(s_0,s_1,t,u) = (s_0+s_1t,t,u). $$ With respect to this isomorphism, the Weil restriction of $X$ with respect to $\phi$ is the basic open affine subset $D(s_0)\subset \mathbb{A}^2_S$. This illustrates one phenomenon of the Weil restriction: although it always transforms affine schemes to affine scheme, it also transforms some (non-affine) quasi-affine schemes to affine schemes.

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