Let $R$ be a torsion-free regular noetherian ring. The Brauer group $Br(R)$ of $R$, defined equivalently (by a theorem of Gabber) as the group of Morita equivalence classes of Azumaya $R$-algebras or as the étale cohomology $H^2(X_{et}, \mathbb{G}_m)$, has the property that it is homotopy invariant: that is, $Br(R) \simeq Br( R[t])$. (I learned this from Auslander and Goldman's article "The Brauer group of a commutative ring." )

Auslander and Goldman's argument is that $Br(R)$ is always a summand of $Br(R[t])$ (because of the natural retraction), so it suffices to show this homotopy invariance property when $R$ is a field $k$ of characteristic zero: this is because Brauer groups of regular domains inject into those of their quotient fields. In this case, one can use Galois descent along $k \to \bar{k}$ together with "Tsen's theorem." (The reason for characteristic zero is simply that Tsen's theorem requires an algebraically closed, rather than separably closed, base field.)

Now this isn't true when the affine line is replaced by a punctured affine line, for instance, $\mathbb{G}_m$ or $\mathbb{A}^1 \setminus \{0, 1\}$: even the Galois descent argument (when $R$ is a field) already breaks down, because the units in the ring of functions on $R[t, t^{-1}]$ is not simply $R^{\times}$, but rather $R^{\times} \oplus \mathbb{Z}$. In particular, if $R = k$ is a field, this will lead to a $H^2( G_k, \mathbb{Z})$ (which is the Pontryagin dual of the Galois group $G_k$) sitting inside the Brauer group of $k[t, t^{-1}]$.

I'm curious if there is a general procedure for calculating the Brauer group of a punctured affine line over a regular ring in terms of the Brauer group of the base ring. (I have in mind something like a localization of $\mathbb{Z}$ as the base.)

  • 1
    $\begingroup$ @VesselinDimitrov: it seems to be typically stated that way, but I don't think it does, since the Galois group is invariant under a purely inseparable extension. I don't know why this restriction to torsion-free rings is therefore necessary. $\endgroup$ Sep 29, 2013 at 20:39
  • 3
    $\begingroup$ Insensitivity of the absolute Galois group to purely inseparable extension is not relevant, as the Galois module is not constant. If $k$ is separably but not algebraically closed of char. $p > 0$ then ${\rm{Br}}(k(t))[p]$ is huge. It suffices (Ch.II, sec.3, Prop.5 in Serre's "Galois cohomology") to make degree-$p$ Galois extensions $L$ of $K=k(t)$ such that the norm $L^{\times}\rightarrow K^{\times}$ is not surjective. For $L$ wild over $\infty$, the norm between those residue fields is $p$-power on $k^{\times}$, so $f$ integral at $\infty$ with $f(\infty)\not\in k^p$ isn't a norm. $\endgroup$
    – Marguax
    Sep 29, 2013 at 21:26
  • 1
    $\begingroup$ @DanielLitt: If I'm not mistaken, if $X$ is regular, the cohomology group $H^2( X_{et}, \mathbb{G}_m)$ injects into the analogous cohomology of the generic point, and so is automatically torsion. $\endgroup$ Sep 30, 2013 at 1:17
  • 1
    $\begingroup$ @Litt: These matters are well-explained in Grothendieck's papers on the Brauer group. $\endgroup$
    – Marguax
    Sep 30, 2013 at 4:32
  • 4
    $\begingroup$ Over a perfect base field $k$, there's an exact sequence due to Faddeev which describes $\textrm{Br }k(t)$ explicitly; this allows you to work out the Brauer group of the affine line with any finite set of points removed. You might look into whether it can be generalised to a base ring, though unfortunately the requirement that the field be perfect is fairly fundamental. $\endgroup$ Sep 30, 2013 at 8:45

1 Answer 1


This is a calculation using the local cohomology sequence for etale cohomology together with Gabber's relatively recent proof of the absolute purity conjecture. Let $% Y=\operatorname{Spec}(R)$, $X=\operatorname{Spec}( R[t]) $, and let $\sigma \subseteq X$ be the image of a section so that $U=\operatorname{Spec}\left( R[t,t^{-1}]\right) $ is its complement. Gabber's recent proof of absolute purity establishes Grothendieck's purity conjecture for the Brauer group which states that $% H_{\sigma }^{3}\left( X,\mathbb{G}_{m}\right) =H^{1}\left( \sigma ,\mathbb{Q}/\mathbb{% Z}\right) $ if $\sigma \subseteq X$ is a regular subscheme of a regular scheme. Moreover $R\rightarrow R[t,t^{-1}]$ has a section. Thus we have an exact sequence coming from the local cohomology of $\mathbb{G}_{m},$ \begin{equation*} 0\rightarrow Br\left( R\right) \rightarrow Br\left( R[t,t^{-1}]\right) \rightarrow H^{1}\left( R,\mathbb{Q}/\mathbb{Z}\right) \rightarrow 0 \end{equation*} since $H^{3}\left( R,\mathbb{G}_{m}\right) =H^{3}\left( R[t],\mathbb{G}_{m}\right) \hookrightarrow H^{3}(R[t,t^{-1}],\mathbb{G}_{m}).$ Note though that this uses Kummer sequences and homotopy invariance for $\mu_n$ and so only applies to torsion in $Br(R[t,t^{-1}])$ of order invertible in $R.$ In fact you can argue directly with $n$ torsion using Kummer sequences and purity for $\mu_n$ cohomology to get this result for $n$ torsion elements if $n$ is a unit in $R$.

Of course if you puncture the line in more than one point, you get multiple copies of $ H^{1}\left( R,\mathbb{Q}/\mathbb{Z}\right)$.

The easiest way to address $n$ torsion when $n$ is not a unit in $R$ is, I think, to look at the split exact sequence on $R_{et}$ $$ 0\rightarrow k_*\mathbb{G}_{m,R[t]} \rightarrow p_*\mathbb{G}_{m,R[t,t^{-1}]} \rightarrow \mathbb{Z} \rightarrow 0 $$ where the last map is the degree map. The Leray spectral sequence for $k$ and $p$ yield a split exact sequence $$ 0\rightarrow Br(R) \rightarrow Br_{sp}(R[t,t^{-1}]) \rightarrow H^1(R,\mathbb{Q}/\mathbb{Z})\rightarrow 0 $$ where $Br_{sp}(R[t,t^{-1}])$ consists of elenents in $Br(R[t,t^{-1}])$ that are split by passing to $R_{\mathfrak{p}}^{sh}[t,t^{-1}]$ as $R_{\mathfrak{p}}^{sh}$ runs through all strict henselizations of $R$ at prime ideals $\mathfrak{p}$.

  • 1
    $\begingroup$ This is interesting, thanks! I think I do need torsion which is not invertible on $R$ though ($R$ for me is something like $\mathbb{Z}[1/2]$). $\endgroup$ Oct 7, 2013 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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