3
$\begingroup$

Let $k$ a separably closed field. Do we have that $k((t))$ is of cohomological dimension one?

$\endgroup$
4
  • $\begingroup$ or at least do we have that any $T$-torsor on $k((t))$ is trivial? As a matter of fact I think that $k((t))$ is of cohomological dimension one only if $k$ is algebraically closed. $\endgroup$
    – prochet
    Jun 8, 2013 at 14:40
  • $\begingroup$ Presumably $T$ is meant to be a torus over $k((t))$, even though this is not said. $\endgroup$
    – user29283
    Jun 8, 2013 at 17:11
  • $\begingroup$ Do you also want a simply connected absolutely simple connected semisimple group over $K = k((t))$ with non-vanishing $H^1$ whenever $k$ is separably but not algebraically closed? If $p := {\rm{char}}(k) > 0$ then there are Galois extensions $K'/K$ of degree $p$, and the norm ${K'}^{\times} \rightarrow K^{\times}$ is not surjective, so there are central division algebras $D$ over $K$ with dimension $p^2$ containing $K'$ as a maximal commutative subalgebra, and $H^1(K, {\rm{SL}}_1(D)) \ne 1$ provided the reduced norm $D^{\times}\rightarrow K^{\times}$ is not surjective. Do you want such $D$? $\endgroup$
    – user29283
    Jun 9, 2013 at 0:31
  • $\begingroup$ An easy case for central division algebras is $p=2$: a smooth conic over $K$ with no $K$-point. Consider $ux^2=y^2+tyz+tz^2$ in $\mathbf{P}^2_K$ for $u\in R^{\times}$ with reduction in $k$ not a square and $t$ a uniformizer. (Allow mixed-characteristic $K$, as in the answer below.) If $[x_0,y_0,z_0]$ is a $K$-point on this conic then WLOG $x_0,y_0,z_0\in R$ with at least one of them a unit. Clearly $z_0$ must be a unit, so WLOG $z_0=1$, and hence $ux_0^2=y_0^2+ty_0+t$. Then ${\rm{ord}}(y_0)={\rm{ord}}(x_0)\le 0$, so we get a contradiction since $u$ isn't residually a square. $\endgroup$
    – user29283
    Jun 9, 2013 at 1:59

2 Answers 2

7
$\begingroup$

No to everything. More generally, consider any complete discrete valuation ring $R$ with uniformizer denoted $t$ and separably closed residue field $k$, and let $K = {\rm{Frac}}(R)$. We allow ${\rm{char}}(K) = 0$, since that case has some interest (e.g., $R$ could be the completion of the maximal unramified extension of the henselized localization of the 2-dimensional regular local ring $\mathbf{Z}_p[\![u,t]\!]$ at the height-1 prime $(u)$; this $R$ has generic characteristic 0, uniformizer $t$, and residue field $\mathbf{F}_p(\!(u)\!)_{\rm{sep}}$.)

Since it is well-known that $K$ has cohomological dimension $\le 1$ when $k$ is algebraically closed, and that fields of cohomological dimension $\le 1$ have vanishing degree-1 Galois cohomology for tori, we just have to show that whenever $k$ is imperfect there is a $K$-torus $T$ for which ${\rm{H}}^1(K,T) \ne 1$.

If $K'/K$ is finite separable then the norm map from $K'$ to $K$ provides an exact sequence of $K$-tori $$1 \rightarrow T \rightarrow {\rm{R}}_{K'/K}({\rm{GL}}_1) \rightarrow {\rm{GL}}_1 \rightarrow 1,$$ so by Hilbert 90 we see that ${\rm{H}}^1(K,T)=1$ if and only if the norm ${K'}^{\times} \rightarrow K^{\times}$ is surjective.

Let $p = {\rm{char}}(k) > 0$ and choose $a \in R^{\times}$ lifting $a_0 \in k-k^p$. The monic polynomial $f = x^p-tx-a \in K[x]$ is irreducible by Gauss' Lemma since it is monic in $R[x]$ with reduction $x^p - a_0 \in k[x]$ that is irreducible. Thus, it is separable if ${\rm{char}}(K)=0$, and it is also separable if ${\rm{char}}(K) = p$ by direct differentiation. Hence, $K' := K[x]/(f)$ is a degree-$p$ separable extension of $K$ and it has valuation ring $R' = R[x]/(f)$ with uniformizer $t$ since this $R'$ is a 1-dimensional noetherian local domain such that $R'/tR' = k[x]/(x^p-a_0)$ is a field. In particular, $K'$ has uniformizer $t$ and residue field $k(a_0^{1/p})$.

(Beware that in the equicharacteristic case, if we fix an isomorphism $R \simeq k[\![t]\!]$ as we may do by the Cohen structure theorem then the residue field of $R'$ does not lift into $R'$ as a $k$-algebra, since $a_0$ is not a $p$th power in $K$ and so cannot be one in $K'$ either, as $K'$ is separable over $K$. That is, although $R' \simeq k(a_0^{1/p})[\![t]\!]$ abstractly as rings -- or even as $\mathbf{F}_p[\![t]\!]$-algebras -- by the Cohen structure theorem, there is no such isomorphism as $k$-algebras or even just as $\mathbf{F}_p(a_0)$-algebras.)

Since ${K'}^{\times} = {R'}^{\times} \times t^{\mathbf{Z}}$ and the norm of $t$ is $t^p$, it is clear that the norm map caries ${K'}^{\times}$ into $R^{\times} \times t^{p\mathbf{Z}}$, so the norm is not surjective. (The norm map on integral units is also generally not surjective, since modulo 1-units the induced map $k(a_0^{1/p})^{\times} \rightarrow k^{\times}$ is readily checked to be the norm map relative to the field extension $k \rightarrow k(a_0^{1/p})$, namely the $p$-power map, and its image $k^p(a_0)^{\times}$ is generally quite thin inside $k^{\times}$, though for $k =\mathbf{F}_p(\!(u)\!)_{\rm{sep}}$ such thinness doesn't occur. But digging one step into the 1-units reveals failure of surjectivity for any $k$ due to the residue field extension being purely inseparable.)

$\endgroup$
0
$\begingroup$

I think the answer to the question about cohomological dim is yes.

In fact, Artin proved in SGA4, exp.xix that, for any complete (or more generally, henselian excellent) equi-char. local domain $A$, $K$ its fraction field, and $k$ the residue field, one has $$cd_l(K)\leq dim(A)+cd_l(k)$$ for every prime $l\neq char(k)$.

In your situation, it's true that the field $k((t))$ may not be a $C_1$ field, nor a field of dim 1 in the sense of Serre(Galois cohomology).

$\endgroup$

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.