Let $f:X\rightarrow Y$ be a smooth morphism of schemes such that all the fibres (for geometric points) are affine spaces. Let $F$ be a coherent sheaf on $X$. Is $R^i_{et}~f_*F=0~~~\forall i>0$? Here "et" in the subscript means the etale higher direct image sheaf.

What if $F$ is the locally constant sheaf $\frac{\mathbb{Z}}{l\mathbb{Z}}$ in the etale topology?

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    $\begingroup$ For coherent sheaves, etale and Zariski cohomology are the same. Thus any non-affine open immersion (e.g. $\mathbb{A}^2\smallsetminus\{(0,0)\}\hookrightarrow\mathbb{A}^2$) gives a counterexample. $\endgroup$ – Laurent Moret-Bailly Oct 28 '18 at 7:13
  • $\begingroup$ I might be reading this question a bit differently from others. I believe that the OP is really asking about affineness for every smooth, surjective morphism whose geometric fibers are affine spaces $\mathbb{A}^n$. When the fibers are allowed to be $\mathbb{G}_m$, this is not true (and I have written some counterexamples in previous MO posts). However, for affine space, I suspect that this is true . . . $\endgroup$ – Jason Starr Oct 28 '18 at 10:10
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    $\begingroup$ No, it is not true. I forgot about my own MathOverflow answer: mathoverflow.net/questions/136017/… $\endgroup$ – Jason Starr Oct 28 '18 at 12:48

It is convenient to formulate the hypotheses as a definition.

Definition 1. A morphism $f:X\to Y$ is an almost affine fibration if $f$ is quasi-compact, quasi-separated, smooth, surjective, and for every point $y$ of $Y$, the fiber $\text{Spec}\ \kappa(y)^{\text{sep}}\times_Y X$ is $\kappa(y)^{\text{sep}}$-isomorphic to an affine space.

Question 2. For every almost affine fibration, for every coherent $\mathcal{O}_X$-module $\mathcal{F}$, does $R^if_*\mathcal{F}$ vanish for every $i>0$.

Of course it suffices to prove this when $Y$ is affine. By limit theorems, even assume that $Y$ is affine and Noetherian. Then also $X$ is Noetherian. Recall the following.

Theorem 3. [Serre's Criterion for Affineness.] A Noetherian scheme $X$ is affine if and only if $H^i(X,\mathcal{I})$ vanishes for every coherent ideal sheaf $\mathcal{I}$ and every integer $i>0$, in which case also $H^i(X,\mathcal{F})$ vanishes for every quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$ and for every integer $i>0$.

Therefore, the original question reduces to the following question.

Question 4. Is every almost affine fibration between Noetherian schemes an affine morphism?

Negative answer, $n\geq 2.$ By my previous MathOverflow answer, the answer is "no" when the fiber dimension is $\geq 2$:

Is an affine fibration over an affine space necessarily trivial?

Notice, the group of automorphisms of affine space $\mathbb{A}^n$ for $n\geq 2$ is enormous.

Arzhantsev, I.; Flenner, H.; Kaliman, S.; Kutzschebauch, F.; Zaidenberg, M.
Flexible varieties and automorphism groups.
Duke Math. J. 162 (2013), no. 4, 767–823.

There might be (further) pathologies caused by nonreduced group schemes in the automorphism group scheme. To avoid such pathologies, it is best to work with $\kappa(y)^{\text{sep}}$ rather than the algebraic closure $\overline{\kappa(y)}$.

Positive result on some étale neighborhood. There is a positive result, at least when the target is integral and after an étale base change.

Lemma 5. Let $f:X\to Y$ be a quasi-compact, smooth, surjective morphism to an integral scheme $Y$ whose separably closed fibers are isomorphic to affine spaces. Then there exists a dominant, quasi-compact, étale morphism $Y'\to Y$ such that $Y'\times_Y X$ is $Y'$-isomorphic to $\mathbb{A}^n_{Y'}$.

Proof. Denote the fraction field of $Y$ by $K$, and denote by $X_\eta$ the fiber product $\text{Spec}\ K \times_Y X$. Denote by $X_{\eta^{\text{sep}}}$ the fiber product $\text{Spec}\ K^{\text{sep}}\times_Y X$. The multiplication map, $$K^{\text{sep}}\otimes_K \mathcal{O}_{X_\eta}(X_\eta) \to \mathcal{O}_{X_\eta^{\text{sep}}}(X_{\eta^{\text{sep}}}),$$ is an isomorphism of $K^\text{sep}$-algebras. By the universal property of affine space, the $K^\text{sep}$-isomorphism from $X_{\eta^{\text{sep}}}$ to $\mathbb{A}^n_{K^{\text{sep}}}$ is given by an ordered $n$-tuple of elements in the coordinate ring of $X_{\eta^{\text{sep}}}$.

By the isomorphism above, each of these elements is a finite $K^{\text{sep}}$-linear combination of elements in the coordinate ring of $X_{\eta}$. There are finitely many $K^{\text{sep}}$-coordinates of these finitely many finite linear combinations. Thus, there exists a finite, separable extension $K'/K$ and a $K'$-morphism $$\text{Spec}\ K' \times_Y X\to \mathbb{A}^n_{K'},$$ whose base change to $K^{\text{sep}}$ is an isomorphism. Since $\text{Spec}\ K^{\text{sep}}$ is faithfully flat and quasi-compact over $\text{Spec}\ K'$, we can use fpqc descent to construct the (unique) inverse isomorphism. Thus, the given $K'$-morphism is already an isomorphism.

By the usual limit theorems, there exists a dominant, quasi-compact, étale morphism $Y'\to Y$ whose extension of function fields is $K'/K$. Denote $Y'\times_Y X$ by $X'$, and denote the generic fiber of $X'\to Y'$ by $X'_{\eta'}$. As above, up to replacing $Y'$ by a dense open subscheme, the ordered $n$-tuple of elements of the coordinate ring of $X'_{\eta'}$ extend to an ordered $n$-tuple of elements of $\mathcal{O}_{X'}(X')$. So there exists a $Y'$-morphism $$g:X'\to \mathbb{A}^n_{Y'},$$ that is an isomorphism on the generic fiber.

The fiber product of $g$ with itself is a closed subscheme $Z$ of $X'\times_{Y'} X'$ that contains the diagonal. The finitely many associated points of $Z$ other than the generic point of the diagonal each maps to a point of $Y'$ other than the generic point. Up to removing from $Y$ the Zariski closures of these points, assume that $g$ is a monomorphism. Similarly, remove from $Y$ the closure of the image of the support of $\Omega_g$. Thus, also assume that $g$ is unramified. Finally, every injective, unramified $\kappa(y)^{\text{sep}}$-morphism, $$\mathbb{A}^n_{\kappa(y)^{\text{sep}}} \to \mathbb{A}^n_{\kappa(y)^{\text{sep}}},$$ is surjective by the Ax-Grothendieck Theorem. Using Grothendieck's form of Zariski's Main Theorem, it follows that $g$ is even finite, hence an isomorphism. QED

By the lemma, it is clear that what goes wrong in examples is that an isomorphism need not specialize to an isomorphism.

Positive answers for $n=1$. The positive results for $n=1$ follow from two straightforward observations.

Lemma 6. For every field $k$, every non-constant $k$-morphism from $\mathbb{A}^1_k$ to $\mathbb{A}^1_k$ is finite.

Proof. This is well-known. The point is that every finitely generated $k$-subalgebra of the polynomial $k$-algebra $k[t]$ is itself a polynomial $k$-algebra generated by some monic polynomial of some degree $d\geq 1$. Then $k[t]$ is generated by $1,t,\dots,t^{d-1}$ over this $k$-subalgebra. QED

Lemma 7. For every field $k$, for every $2$-pointed affine curve $(C,p_0,p_1)$ that is abstractly $k$-isomorphic to $(\mathbb{A}^1_k,0,1)$, for every triple $(g,h_0,h_1)$ of an element $g\in \mathcal{O}_C(C)$ that is separable, i.e., $dg$ is not identically zero in $\Omega_{C/k}(C)$, and a pair of elements, $h_i\in \mathcal{O}_C(C\setminus\{p_i\}),$ $i=0,1$, such that $g(p_0)=0$, such that $g(p_1)=1$, such that $gh_0=1$ on $C\setminus\{p_0\}$, and such that $(g-1)h_1=1$ on $C\setminus\{p_1\}$, then $g$ defines the unique $k$-isomorphism from $(C,p_0,p_1)$ to $(\mathbb{A}^1_k,0,1)$.

Proof. By hypothesis, there exists a $k$-isomorphism from $(C,p_0,p_1)$ to $(\mathbb{A}^1_k,0,1)$. Thus, assume that $(C,p_0,p_1)$ equals $(\mathbb{A}^1_k,0,1)$, and let $t$ denote the usual coordinate. Since $g$ is invertible on the complement of $0$, the element $g$ must equal $ct^d$ for some $c\in k^\times$ and some integer $d\geq 0$. Since $g-1$ is invertible on the complement of $1$, $ct^d-1$ equals $c(t-1)^e$ for some integer $e\geq 0$. Thus, $c$ equals $1$ and either $d=e=1$, or $d=e=p^r$ with $p=\text{char}(k)$ and $r\geq 1$ an integer. In the second case, the differential of $g$ is identically zero. Thus, $g$ equals $t$. QED

Proposition 8. Let $R$ be a DVR, and let $X$ be a quasi-compact, quasi-separated, smooth $R$-scheme together with two disjoint $R$-sections, $\sigma_0$ and $\sigma_1$ such that the fibers over the separable closures of $\text{Frac}(R)$ and $R/\mathfrak{m}$ are isomorphic to $(\mathbb{A}^1,0,1)$. Then $X$ is $R$-isomorphic to $\mathbb{A}^1_R$ with its zero section and one section.

Proof. The uniqueness of $g$ from Lemma 7 allows to descend the isomorphism from $\mathbb{A}^1_{\text{Frac}(R)^{\text{sep}}}$ to an isomorphism of generic fibers, $$g:X_{\text{Frac}(R)}\to \mathbb{A}^1_{\text{Frac}(R)}.$$ For the open immersion $$u:\mathbb{A}^1_{\text{Frac}(R)} \hookrightarrow\mathbb{P}^1_R,$$ denote by $\Gamma$ the closure of the graph of $u\circ g$ in $X\times_{\text{Spec}\ R}\mathbb{P}^1_R.$ The projection morphism from $\Gamma$ to $X$ is projective and birational, even an isomorphism over the maximal open subscheme $X_g$ of $X$ on which $g$ is regular.

By construction, $X_g$ contains the generic fiber. By the valuative criterion of properness, $X_g$ intersects the closed fiber. Since the total degree of the projection $\Gamma \to \mathbb{P}^1_R$ equals $1$ on the generic fiber, it also equals $1$ on the closed fiber. Thus, there is at most one irreducible component of the closed fiber of $\Gamma$ that is not contracted by the morphism to $\mathbb{P}^1_R$. In particular, $g$ is regular along $\sigma$, where $\sigma$ is either the zero section or the one section. Denote by $\tau$ the corresponding section of $\mathbb{A}^1_R$. The derivative of $g$ is an $R$-module homomorphism from $\sigma^*\Omega_{X/R}$ to $\tau^*\Omega_{\mathbb{A}^1_R/R}$. By Lemma 7, this is an isomorphism of $R$-modules. Thus, $g$ restricts to a nonconstant element on an open neighborhood of the closed fiber of $X$ that intersects $\sigma$.

Since the strict transform of the closed fiber of $X$ is the unique irreducible component of the closed fiber of $\Gamma$ that is not contracted, in fact the morphism $\Gamma \to X$ extracts no divisors. Thus, $g$ is everywhere regular on $X$. Since the restriction of $g$ to the closed fiber is nonconstant of degree $1$, the morphism $g$ is an isomorphism on closed fibers. Thus, $g$ is an isomorphism from the $R$-scheme $X$ to the $R$-scheme $\mathbb{A}^1_R$. QED

Proposition 9. For every almost affine fibration of relative dimension $1$, if $Y$ is normal, then the morphism is affine. Moreover, after an étale, surjective base change of $Y$, the morphism is isomorphic to projection from $\mathbb{A}^1_Y$ to $Y$.

Proof. For every point $y$ of $Y$, since $f$ is smooth, after a quasi-compact, dominant, étale morphism $Y'\to Y$ whose image contains $y$, there exist two disjoint sections of $f$. Call these the zero section and the one section. The identity morphism is the unique automorphism of $\mathbb{A}^1$ that preserves $0$ and $1$. By Lemma 5, this defines a morphism $g$ to $\mathbb{A}^1_{Y'}$ on a dense open subset of $X'=Y'\times_Y X$. Moreover, after deleting the closed image of the zero section, resp. of the one section, also $h_0=1/g$, resp. $h_1=1/(g-1)$, is regular on a dense open subset.

Since $Y'$ is normal and since $f$ is smooth, also $X'$ is normal. By the proposition, $g$ and $h_0$, resp. $h_1$, are regular over every codimension $1$ point of $Y$, resp. over every codimension $1$ point of $Y$ after deleting the zero section, resp. the one section. Since $X'$ is normal, a rational function to $\mathbb{A}^1_R$ that is regular at codimension $1$ points is everywhere regular. Thus, $g$ and $h_0$, resp. $h_1$, are regular on all of $X'$, resp. on the complement of the zero section, resp. the complement of the one section. By Lemma 7, the scheme $X'$ is $Y'$-isomorphic to $\mathbb{A}^1_{Y'}$. In particular, $X'\to Y'$ is an affine morphism. Since the property of affineness of a morphism can be checked after faithfully flat base change, it follows that the original morphism is also affine. QED


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