Examples of Galois-invariant central simple algebras which aren't base change? Suppose $L/K$ is a Galois extension of number fields, with Galois group $G_{L/K}$.  Write $\mathrm{Br}(L)^{G_{L/K}}$ for the subgroup of central simple algebras $A/L$ which are Galois-invariant; equivalently, these are the algebras such that for $v$ a place of $K$ and $w$ a place of $L$ with $w|v$, the local invariants $\mathrm{inv}_w(A)$ only depend on $v$.  The Hochschild-Serre spectral sequence yields an exact sequence
$0\to \mathrm{Br}(L/K) \to \mathrm{Br}(K) \to \mathrm{Br}(L)^{G_{L/K}}\to H^3(G_{L/K},L^{\times}),$
where the third arrow is represented by the base change map $A\mapsto A\otimes_{K} L$, and the image of the fourth arrow is the "obstruction" preventing a Galois invariant algebra from arising via base change.  Now, the $H^3$ appearing here is often zero; for example, it vanishes if all the Sylow subgroups of $G_{L/K}$ are cyclic.  My questions:


*

*In the case $K=\mathbf{Q}$ (say), what is the simplest example of a Galois extension $L/\mathbf{Q}$ and a Galois-invariant central simple algebra over $L$ which is not a base change from any proper subfield? Does $G_{L/\mathbf{Q}}=A_4$ work?

*Is there a simple way to compute the obstruction map into $H^3$?  It seems like the answer to this must naively be "no", since it cannot be as simple as "do such and such with the local invariants".

*How much of the above goes through without assuming $L/K$ is Galois?
 A: [big edit]
(1) Let $L/K$ be as above. Take any non-identity element $\sigma \in G_{L/K}$, and let $F=L^{<\sigma>}$ be the fixed field of the cyclic subgroup generated by $\sigma$. By your comment above about the vanishing of $H^3$, every galois invariant CSA of $L$ is a base change of a CSA of $F$. Hence, there are no galois invariant CSA that are not a base change from any proper subfield.
But, if you fix $L$ and $K$, then there can be galois invariant CSA's of $L$ that are not a base change from $K$. Since the smallest non-cyclic group is $C_2\times C_2$, we would like to search for examples with such a galois group.
For $K=\mathbb{Q}$, probably, the simplest example is:
$$L=\mathbb{Q}(\sqrt{-3},\sqrt{13}),\ (43)=P_1\cdot P_2\cdot P_3\cdot P_4,$$
$$u=\frac{1}{4}P_1+\frac{1}{4}P_2+\frac{1}{4}P_3+\frac{1}{4}P_4 \in \bigoplus_v Br(L_v)$$
I will prove below that this element of the Brauer group is not a base change, and that in fact a similar construction can be given for any extension.
(2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, there is a small paragraph devoted to $H^3(G_{L/K},L^\times)$:

"$H^3(G,L^\times)$ is cyclic of order $n/n_0$, the global degree divided by the lowest common multiple of local degrees, generated by $\delta u_{L/K}$ ($\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), the "Teichmuller 3-class." ..."

Therefore, assume $n_0 < n$ for our galois extension $L/K$. Let $v_0$ be an unramified finite place of $K$ that splits completely (exists by Chebotarev's theorem). We can construct an element of the Brauer group similar to the one above:
$$u := \sum_{w|v_0} \frac{1}{n} w$$
This is in fact an element of $Br(L)$ since the number of $w|v_0$ is $n$, so that $n\frac{1}{n} = 1 \in \mathbb{Z}$.

Proposition. For any prime $p$ that divides $\frac{n}{n_0}$, $\frac{n}{pn_0}\cdot u$ is not a base change.

From which we immediately get:

Corollary. The map $Br(L)^{G_{L/K}}\rightarrow H^3(L^\times)$ is onto.

Proof of proposition. Assume that $\frac{n}{pn_0}\cdot u$ is the base change of some $u'$, i.e.
$$u' = \sum_v n_v v \mapsto \sum_v \sum_{w|v} [L_w : K_v] n_v w = \frac{n}{pn_0}\cdot u$$
Since for any $w|v_0$: $[L_w : K_{v_0}] = 1$, we must have $n_{v_0} = \frac{n}{pn_0}\cdot \frac{1}{n} = \frac{1}{pn_0}$. And since $\sum_v n_v \in \mathbb{Z}$, at least one other place $v_1$ has
$$v_p(n_{v_1}) \le v_p(n_{v_0}) = v_p(\frac{1}{pn_0}) < 0$$
Where $v_p$ is the usual $p$-adic valuation. So, using that $n_0$ is the lcm of local degrees:
$$v_p([L_{v_1}:K_{v_1}] n_{v_1}) \le v_p(n_0) + v_p(\frac{1}{pn_0}) = -1$$
Contradicting the zero coefficient of any $w|v_1$ in $u$.
A small computation shows that for the extension $\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$, $n_0=2$, and that indeed this is the smallest (discriminant-wise) $C_2\times C_2$ such extension.
A: I don't think that you have an obstruction on the level of Brauer groups. If you have a Galois-stable element over $L$, then you can choose some values of local invariants for its lift over $K$ such that their sum would be $0$ (in $\mathbb{Q}/\mathbb{Z}$). To this end you should lift your local invariants to elements of $\mathbb{Q}$ whose sum is $0$, and divide them by the degrees of the corresponding local extensions.
On the other hand, it seems that often the degree of any of the lifts (over $K$) is greater than the one of the original element (over $L$); hence on the level of algebras you can only lift a certain matrix algerbra over your original central divisible algebra.
