Let $F$ be a local field of characteristic zero (for simplicity), $\overline{F}$ an algebraic closure of $F$ and $L/F$ a fixed finite Galois extension. If $G$ is a linear algebraic group defined over $F$, then the Galois cohomology group $H^1(F,G)$ can be defined as a direct limit of $H^1(K/F,G)$, where $K$ runs through finite Galois subextensions of $\overline{F}$.


Now the question is: under what conditions is this direct limit just $H^1(L/F,G)$?


I guess there might be restrictions on both $L$ and $G$. 



Galois cohomology of linear groups over local fields
As Hunter and Sean noted, since the inflation map ${\rm{H}}^1(L/F,G(L)) \rightarrow {\rm{H}}^1(F,G)$ is
injective and ${\rm{H}}^1(F,G)$ is always finite (Borel-Serre), such an $L$ always exists. Below we give an explicit sufficient condition on $L$ (often satisfied) when $G$ is connected. (One could probably do better with a closer consideration of the Tate local duality aspects of the argument. I am lazy at this step.) This rests on something deeper than the Borel-Serre result: the Kneser-Bruhat-Tits theorem on vanishing of degree-1 Galois cohomology for simply connected semisimple groups over non-archimedean local fields. 


First we set up some notation.
Let $U = \mathcal{R}\_u(G)$ denote the unipotent radical of $G$ (this is a good notion over $F$ since $F$ is perfect), and let $G' = G/U$ denote the maximal reductive quotient. The identity component $Z'$ of the center of $G'$ is an $F$-torus, and the derived group $\mathcal{D}(G')$ is a semisimple group, say with simply connected central cover $\mathcal{G} \twoheadrightarrow \mathcal{D}(G')$. The preimage $\mu$ of the central
subgroup $Z' \cap \mathcal{D}(G')$ is a finite $F$-group of multiplicative type. 
(It is the kernel of the central covering map if $Z' = 1$.) 


Proposition: Assume $G$ is connected and use notation as above. 
Let $F'/F$ be a finite Galois splitting field for $G'$ (thus for $Z'$ and dual of
$\mu$). If $L/F$ is a finite Galois extension containing $F'$ with $[L:F']$ divisible by the order of $\mu$ then ${\rm{H}}^1(F,G) = {\rm{H}}^1(L/F,G(L))$.


In particular, if $G$ is a split connected reductive $F$-group then $L/F$ works provided that $[L:F]$ is divisible by the order of the center of the simply connected
central cover of $\mathcal{D}(G)$. 


Remark: If $T$ is a maximal $F$-torus in $G$ then it maps isomorphically onto one for $G'$, so could take $F'/F$ to be splitting field for $T$. 


Proof: Since $F$ has characteristic 0, the quotient map $G \twoheadrightarrow G'$ admits a section $\sigma$ over $F$, which is to say there exists a connected reductive $F$-subgroup $H \subseteq G$ such that $H \ltimes U \simeq G$ via multiplication (this is a so-called Levi $F$-subgroup of $G$); beware that over any algebraically closed field $k$ with nonzero characteristic Levi subgroups can fail to exist. (A basic natural counterexample is, loosely speaking, ${\rm{SL}} \_2(W \_2(k))$ as a $k$-group, where $W \_2$ denotes length-2 Witt vectors; see Appendix A.6 in the book "Pseudo-reductive groups".)


Using $\sigma$ (or $H$), the natural restriction map ${\rm{H}}^1(F,G) \rightarrow {\rm{H}}^1(F,G')$ is surjective. It is also injective. Indeed, a standard twisting argument (as explained in Serre's book on Galois cohomology) identifies fibers with ${\rm{H}}^1(F,U')$ for various $F$-forms $U'$ of $U$. But since the ground field is perfect, every smooth connected unipotent group is split (i.e., admits a composition series whose successive quotients are $\mathbf{G}\_a$) and hence has trivial ${\rm{H}}^1$. Thus, we get the asserted bijectivity. So far we have not used anything about $F$ other than that it has characteristic 0. 


We likewise have ${\rm{H}}^1(E/F,U'(E)) = 1$ for any smooth connected unipotent $F$-group $U'$ and any Galois extension $E/F$, so the same argument gives that ${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(E/F,G'(E))$ is bijective for any Galois extension $E/F$. Thus, the injective inflation map 
$${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(F,G)$$
is bijective if and only if the same holds for $G'$ in place of $G$. 


Consider the central extension structure 
$$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$
over $F$, where the second map uses multiplication. By Kneser-Bruhat-Tits, 
we have an exact sequence of pointed sets
$${\rm{H}}^1(F,Z') \rightarrow {\rm{H}}^1(F,G') \rightarrow {\rm{H}}^2(F,\mu)$$
and similarly over $L$, with a commutative diagram using restriction maps. 


By local class field theory, the restriction map from ${\rm{H}}^2(F',\mu)$ to 
${\rm{H}}^2(L,\mu)$ vanishes, so likewise for ${\rm{H}}^2(F,\mu) \rightarrow {\rm{H}}^2(L,\mu)$. (This is weak; by translating restriction through Tate local duality we can surely give a better sufficient "lower bound" for such an $L$.) For any such $L$, it follows that restriction from ${\rm{H}}^1(F,G')$ to ${\rm{H}}^1(L,G')$ lands in the image of ${\rm{H}}^1(L,Z')$. But this vanishes since the $F$-torus $Z'$ is even $F'$-split, let
alone $L$-split. Thus, ${\rm{H}}^1(F,G') \rightarrow {\rm{H}}^1(L,G')$ vanishes, which is to say that $L$ "works" for $G'$, so the same holds for $G$. 
QED


