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

## 2 Answers

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.)

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).

providedthe reduced norm $D^{\times}\rightarrow K^{\times}$ is not surjective. Do you want such $D$? $\endgroup$smoothconic 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. $\endgroup$