Extending commuting automorphisms of a field I have a field $L$ and two field automorphisms $f, g: L \to L$ such that $f \circ g = g \circ f$. Under what conditions can they be extended to automorphisms of an extension $K \mid L$ that still commute? In particular they can certainly be extended to automorphisms of the algebraic clousure $K = \overline L$, but can they be extended to a pair of commuting automorphisms?
 A: Reformulation in terms of group cohomology of Galois groups.  Assume that $f$ and $g$ both have finite order dividing  the positive integer $n.$  The the fixed subfield $E=L^{f,g}$ of $f$ and $g$ gives a finite, Galois extension $L/E.$  The two commuting elements $f$ and $g$ define a surjective group homomorphism, $$\rho_{f,g,n}:\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z} \twoheadrightarrow \text{Gal}(L/E).$$  
For every finite, separable extension $K/L$, since also $K/E$ is finite and separable, there exists a minimal Galois extension $K'/E$ that contains $K/E$ as a subextension.  By the Fundamental Theorem of Galois Theory, $K'/L$ is a Galois extension, the subgroup $\text{Gal}(K'/K)$ is a normal subgroup, $\text{Gal}(K'/L)\trianglelefteq \text{Gal}(K'/E),$ and the quotient by this subgroup is canonically isomorphic to $\text{Gal}(L/E).$  To summarize, there is a normal extension of groups, $$\Sigma_{K'/L/E}: \ \ \ 1 \to \text{Gal}(K'/L)\to \text{Gal}(K'/E) \to \text{Gal}(L/E) \to 1.$$  In particular, there are lifts $\widetilde{f},\widetilde{g}\in \text{Gal}(K'/E)$ mapping to $f,g\in \text{Gal}(L/E).$  
Up to increasing $n$, assume that the order of $\text{Gal}(K'/E)$  divides $n$.  Then the pullback of $\Sigma_{K'/L/E}$ by $\rho_{f,g,n}$ is a normal extension of groups, $$\Sigma_{K'/L,f,g,n}: \ \ \ \ 1 \to \text{Gal}(K'/L) \to G \to \mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z} \to 0.$$ 
Proposition. There exist lifts $\widetilde{f}$ and $\widetilde{g}$ of $f$ and $g$ in $\text{Gal}(K'/E)$ that commute if and only if the normal extension $\Sigma_{K'/L,f,g,n}$ is a semidirect product, i.e., if and only if the extension is split.  If $\text{Gal}(K'/L)$ is Abelian, the extension $\Sigma_{K'/L,f,g,n}$ is an element in the group cohomology $H^2(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},\text{Gal}(K'/L)).$
Proof.  If there are lifts that commute, then the subgroup generated by $\widetilde{f}$ and $\widetilde{g}$ gives a section of $\Sigma_{K'/L,f,g,n}.$  Conversely, if there is a section, then the images in $\text{Gal}(K'/E)$ of the standard generators of $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ commute. QED
The "elementary obstruction" to existence of simultaneous fixed points. The extension $\Sigma_{K'/L,f,g,n}$ need not always split.  The simplest such example is any Galois extension $K'/E$ with Galois group equal to the order $8$ quaternion group $Q$.  Take $L$ to be the fixed field of the center of $Q$.  The corresponding normal extension of groups is not a semidirect product.  
There are also many examples arising in algebraic geometry from the problem of trying to find simultaneous fixed points of two commuting automorphisms of a $k$-scheme.  
Setup.  Let $k$ be a field of characteristic prime to $n$ that contains a primitive $n^{\text{th}}$ root of unity, $\zeta.$  Let $P$ be a projective, integerl $k$-scheme $P$ whose Picard group is cyclic generated by an ample invertible sheaf $\mathcal{O}(1).$  Let $f,g:P\to P$ be $k$-automorphisms of finite order dividing $n$ that commute with each other.
Problem.  Does there exists a $k$-point of $P$ that is simultaneously fixed by $f$ and by $g?$
N.B.  For either $f$ or $g$ individually, there are many fixed point theorems that insure the existence of a fixed point.  For instance, if $k$ is algebraically closed and $P$ is smooth and separably rationally connected, then there exists a fixed point of $f$, resp. there exists a fixed point of $g$.  However, there may exist no simultaneous fixed point, and one obstruction arises from failure of commutativity of lifts of $f$ and $g$. 
Heisenberg group examples.  Let $P$ be $\mathbb{P}^{n-1}_k = \text{Proj}\ k[t_0,\dots,t_{n-1}].$  Consider the following $k$-automorphisms, $$f:\mathbb{P}^{n-1}_k \to \mathbb{P}^{n-1}_k, \ \ [t_0,t_1,\dots,t_{n-2},t_{n-1}] = [t_{1},t_2,\dots,t_{n-1},t_0],$$
$$g:\mathbb{P}^{n-1}_k \to \mathbb{P}^{n-1}_k, \ \   
[t_0,t_1,\dots,t_{n-1}] \mapsto [\zeta^0 t_0, \zeta^1t_1,\dots,\zeta^{n-1}t_{n-1}].$$  The fixed points of $g$ are the points $p_i = [0,\dots,0,1,0,\dots,0],$ $i=0,\dots,n-1,$ with precisely one nonzero coordinate.  The fixed points of $f$ are the points $[\zeta^{0\cdot a},\zeta^{1\cdot a}, \dots, \zeta^{(n-1)\cdot a}]$ for $\overline{a}\in \mathbb{Z}/n\mathbb{Z}.$  There is no simultaneous fixed point of $f$ and $g$.
Cyclic Galois extensions.
For each integer $e$, the automorphisms $f$ and $g$ "linearize" to a $k$-linear action of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ on the $k$-vector space $$V_e=H^0(P,\mathcal{O}(n\cdot e)).$$  Since the characteristic is prime to $n,$ every such $k$-representation is a direct sum of $1$-dimensional $k$-subrepresentations.  Let $L=\text{span}(\sigma)$ be a $1$-dimensional $k$-subrepresentation.  Let $q:Y_\sigma\to P$ be the cyclic $n$-sheeted cover on which there is a universal section $s$ of $f^*\mathcal{O}(e)$ with $s^n$ equal to the pullback of $\sigma$.  Over the dense Zariski open $D(\sigma)\subset P$, the morphism $q$ is a cyclic Galois cover with Galois group $\mu_n(k)$ acting by scaling $s$.  Thus, every lift $\widetilde{f},$ resp. $\widetilde{g},$ of $f,$ resp. $g,$ is unique up to scaling by an element $\phi,$ resp. $\gamma,$ in $\mu_n(k).$
Definition. The elementary obstruction to the existence of a simultaneous fixed point is the commutator $o_{f,g,\sigma}=\widetilde{f}\circ \widetilde{g}\circ \widetilde{f}^{-1}\circ \widetilde{g}^{-1}$ in $\mu_n(k).$  Since $\phi\cdot \gamma\cdot \phi^{-1}\cdot \gamma^{-1}$ equals $1$ in the Abelian group $\mu_n(k)$, the elementary obstruction is independent of the choice of lifts.
Lemma.  If $P$ is smooth and if there exists a $k$-point of $P$ that is simultaneously fixed by $g$ and by $g$, then the elementary obstruction equals the identity element $1$ in $\mu_n(k).$
Proof.  By Lang-Nishimura, existence of a fixed points is a birational invariant.  Thus, up to blowing up and taking strict transforms, assume that there exists a simultaneous fixed point $x$ that is contained in the open $D(\sigma)$.  The fiber of $q$ over $x$ is a torsor for $\mu_n(k)$.  Thus, the lifts $\widetilde{f}$ and $\widetilde{g}$ restrict on this fiber to scaling by elements of $\mu_n(k)$.  Up to scaling by appropriate elements $\phi$ and $\gamma$, assume that these are the identity.  Then the commutator equals $1$ on this fiber.  Since the commutator is scaling by a constant element $\mu_n(k),$ it follows that the commutator equals $1$. QED
As above, this obstruction is naturally an element in the group cohomology $H^2(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},\mu_n(k)).$  In fact, this obstruction depends only on the integer $e$, not on the nonzero element $\sigma,$ i.e., $o_{f,g,\sigma} = o_{f,g,e}.$  Moreover, $o_{f,g,e}$ equals $(o_{f,g})^e$, where $o_{f,g}$ equals $o_{f,g,1}.$
Computation of the elementary obstruction in the Heisenberg examples.
For $P=\mathbb{P}^{n-1}_k,$ for the invariant section $\sigma = t_0^n + \dots + t_{n-1}^n,$ over each basic open $D_+(t_i)$, the inverse image of $Y_\sigma$ is equal to the smooth, integral, affine scheme, $$\text{Spec}\ k[s,(t_j/t_i)]_{j\neq i} /\langle s^n - (t_0^n+\dots+t_{n-1}^n)/t_i^n \rangle.$$  The automorphism $f$ lifts to an automorphism, $$(s,(t_j/t_i))_{j\neq i} \mapsto (s(t_{i+1}/t_i)^{-1}, (t_{j+1}/t_i) (t_{i+1}/t_i)^{-1})_{j\neq i},$$ and this is unique up to scaling $s$ by a scalar $\phi.$  The automorphism $g$ lifts to an automorphism, $$(s,(t_j/t_i))\mapsto (s,\zeta^{j-i}(t_j/t_i)),$$ and this is unique up to scaling $s$ by a scalar $\gamma.$  For the lifts just given, the commutator $o_{f,g,\sigma} = \widetilde{f}\circ \widetilde{g}\circ \widetilde{f}^{-1}\circ \widetilde{g}^{-1}$ equals $\zeta^{-1}.$  Thus, this nontrivial obstruction "explains" the failure of simultaneous fixed points of $f$ and $g.$
Having said this much, I feel compelled to explain one family of positive examples.
Theorem[A. J. de Jong-Xuhua He-S.]  Let $k$ be algebraically closed, and let $P$ be a smooth, projective $k$-scheme that is a specialization of a "rationally simply connected" variety of Picard rank $1,$ e.g., a projective homogeneous space of Picard rank $1$ (like a Grassmannian) or a complete intersection in projective space with both $\text{c}_1(T_P)$ and $\text{c}_1(T_P)^2 - 2c_2(T_P)$ positive ($2$-Fano complete intersection).  If the elementary obstruction $o_{f,g}$ vanishes, then there exists a simultaneous fixed point.
