Are Z-torsors on A1 trivial?

[felixwellen]

In other words: Do we have $H^1(\mathbb A^1,\mathbb Z)=0$ ? Using reasoning from 2.2 in the A1-homotopy draft, this would also imply that $B\mathbb Z=S^1$ is $\mathbb A^1$-local.

[iblech]

(Note: $\mathbb{Z}$ is not in any sense quasicoherent -- it is not even an $R$-module. Else vanishing would be trivial in view of Serre's criterion for affine schemes.)

Do we have the vanishing in classical algebraic geometry? Yes, for the trivial reason that the underlying topological space of an integral scheme always has vanishing higher singular cohomology.

[felixwellen]

It is not completely clear to me, that $H^1(\mathbb A^1,\mathbb Z)=|\mathbb A^1\to B\mathbb Z|_0$ is actually the singular cohomology. I guess what we are trying to show internally will actually use arguments which should also be of use when showing that $H^n(X,\mathbb Z)$ is the singular cohomology of a scheme $X$. And I guess the following "topological" argument should actually work if we assume overtness of $\mathbb A^1$: Let's say we have a $\mathbb Z$-torsor given as a map $t:T\to\mathbb A^1$. So the fibers of $t$ are all merely $\mathbb Z$. Then by Zariski-local choice, we get a cover $U_1,\dots,U_n$ and local sections $s_i$. Let's say without loss of generality, that $0\in U_1$, which we can do, because we want to prove the proposition that $t$ merely has a section. We know that all $U_1\cap U_i$ are irreducible, since $\mathbb A^1$ is irreducible and open subsets of irreducibles are irreducible (written down in this draft). A map from an irreducible scheme to $\mathbb Z$ is weakly constant, since pointed irreducibles are connected. Now if we want to define a section $s:\mathbb A^1\to T$ of $t$ for some $x:\mathbb A^1$, we may proceed by defining a value for each $i$ with $x\in U_i$ and then show that those values are pairwise equal. So for $x\in U_i$ we know by irreducibility of $\mathbb A^1$ that $U_1\cap U_i$ is non-empty and by overtness of $\mathbb A^1$ that it is inhabited. So for $y\in U_1\cap U_i$ we get a difference $k:\equiv s_1(y)-s_i(y)$ which we can add to $s_i$ to correct it to some $\widetilde{s_i}$. So $\widetilde{s_i}(x)$ should be the same value for all $i$ with $x \in U_i$ and we can use it to define $s(x)$.