[big edit]

Let's

I will prove below that this element of the CSA corresponding to $u$ Brauer group is indeed not a base changefrom $\mathbb{Q}$.

Proposition. Let $L/K$ be , and that in fact a galois similar construction can be given for any extensionwith galois group $C_2\times C_2$.

(2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, such that no ramified finite place there is a small paragraph devoted to $H^3(G_{L/K},L^\times)$:

"$H^3(G,L^\times)$ is cyclic of order $K$ has n/n_0$, the global degree divided by the lowest common multiple of local degrees, generated by$g=1$\delta u_{L/K}$ (usual notation satisfying $n=efg$). Let $k$ be a positive integer$\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), and the "Teichmuller 3-class." ..."

Therefore, assume $p_i$, n_0 < n$for our galois extension$1\le i\le 2^k$, L/K$. Let $v_0$ be an unramified finite places place of $K$ that split splits completely $p_i=P_1^i\cdot P_2^i\cdot P_3^i\cdot P_4^i$(exists by Chebotarev's theorem). Then We can construct an element of the CSA over $L$ corresponding Brauer group similar to the one above:$$u=\sum_{i=1}^{2^k} u := \frac{1}{2^{k+2}}P_1^i+\frac{1}{2^{k+2}}P_2^i+\frac{1}{2^{k+2}}P_3^i+\frac{1}{2^{k+2}}P_4^i sum_{w|v_0} \frac{1}{n} w$$

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 \bigoplus_w Br(L_w)$$in \mathbb{Z}. Proposition. For any prime p that divides \frac{n}{n_0}, \frac{n}{pn_0}\cdot u is not a base changeof a CSA over . From which we immediately get: Corollary. The map K.Br(L)^{G_{L/K}}\rightarrow H^3(L^\times) is onto. Proof of proposition. Assume that \frac{n}{pn_0}\cdot u is the base change of some u'=\sum_v n_v v, u', i.e.$$u' = \sum_v n_v v \mapsto \sum_v \sum_{w|v} (\frac{4}{g_v} [L_w : K_v] n_v )w = \frac{n}{pn_0}\cdot u$$Since for each of the p_i: f=1, any e=1, w|v_0: g=4, [L_w : K_{v_0}] = 1, we must have n_{p_i}=\frac{1}{2^{k+2}}. We know that n_{v_0} = \frac{n}{pn_0}\cdot \frac{1}{n} = \frac{1}{pn_0}. And since \sum_v n_v=0 (n_v \in \mathbb{Q}/\mathbb{Z}), and the contribution from the p_i's is 2^k\frac{1}{2^{k+2}}=\frac{1}{4}, hence \mathbb{Z}, at least one other place has its n_v's denominator divisible by 4. Since Br(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}, this other place must be a finite place, q. If q is unramified, then by basic number theory g_q\ne 1 (L is a composite of two quadratic extensions, L=K(\sqrt{a},\sqrt{b}), so at least one of a, b, ab is a square modulo v_1 hasq). If v_p(n_{v_1}) \le v_p(n_{v_0}) = v_p(\frac{1}{pn_0}) < 0$$ Where$q$v_p$ is ramified, then by the hypothesis, usual $g_q\ne 1$p$-adic valuation. This shows that$g_q \in \{2,4\}$. HenceSo, the denominator of$(\frac{4}{g_q})n_q$must be divisible by$2$, contradicting the fact using that for any$w|q$,$n_w=0$in$u$. The smallest example, discriminant-wise (1521), of such an extension of$\mathbb{Q}$is the one cited above.[EDIT] (2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, there is a small paragraph devoted to$H^3(G_{L/K},L^\times)$: "$H^3(G_{L/K},L^\times)$n_0$ is cyclic of order $n/n_0$, the global degree divided by the lowest common multiple lcm of local degrees, generated by :$\delta u_{L/K}$ ($\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), the "Teichmuller 3-class." ..."

So in $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 case of a C_2\times C_2 extension zero coefficient of number fields, satisfying the any g\ne 1 hypothesis as w|v_1 in the proposition, we see that the fourth arrow is onto. On the other hand, any C_2\times C_2 extension u. A small computation shows that doesn't satisfy for the extension g\ne 1 hypothesis has \mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}, n_0=4, n_0=2, and hence H^3(L^\times)=0. This settles all that indeed this is the smallest (discriminant-wise) C_2\times C_2 extensions. As for A_4, or any other such extension, it seems some calculations with the fundamental class and the associated Teichmuller 3-class are in order. But, since the only reason I can quote the above is that I "skipped to the end" while reading the chapter, I will let someone else go for it. 5 added Tate reference; edited body; added 1 characters in body (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 a not a base change from any proper subfield. [EDIT] (2) I won't pretend to understand anything about the last arrow. The only thing I will say is that Near the above propositionend of Tate's "Global Class Field Theory" section in Cassels-Frohlich, proven with there is a bit small paragraph devoted to H^3(G_{L/K},L^\times): "H^3(G_{L/K},L^\times) is cyclic of over-generalisation (if only order n/n_0, the example is desired)global degree divided by the lowest common multiple of local degrees, gives generated by \delta u_{L/K} (\delta:H^2(C_L)\rightarrow H^3(L^{\times})), the "Teichmuller 3-class." ..." So in the case of a lot C_2\times C_2 extension of CSA's that aren't base change. It is also interesting to note that if we remove number fields, satisfying the g \ne g\ne 1 hypothesis as in the proposition, then I think we see that there the fourth arrow is once again no obstructiononto. On the other hand, even though any C_2\times C_2 extension that doesn't satisfy the H^3 g\ne 1 hypothesis has n_0=4, and hence H^3(L^\times)=0. This settles all C_2\times C_2 extensions. As for A_4, or any other extension, it seems some calculations with the fundamental class and the associated Teichmuller 3-class are in order. But, since the only reason I can quote the above is not trivialthat I "skipped to the end" while reading the chapter, I will let someone else go for it. 4 added 3 characters in body; added 3 characters in body (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 a not 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 \oplus_v bigoplus_v Br(L_v)$$Let's prove that the CSA corresponding to u is indeed not a base change from \mathbb{Q}. Proposition. Let L/K be a galois extension with galois group C_2\times C_2, such that no ramified finite place of K has g=1 (usual notation satisfying n=efg). Let k be a positive integer, and p_i, for 1\le i\le 2^k, be unramified finite places that split completely p_i=P_1^i\cdot P_2^i\cdot P_3^i\cdot P_4^i. Then the CSA over L corresponding to$$u=\sum_{i=1}^{2^k} \frac{1}{2^{k+2}}P_1^i+\frac{1}{2^{k+2}}P_2^i+\frac{1}{2^{k+2}}P_3^i+\frac{1}{2^{k+2}}P_4^i \in \oplus_w bigoplus_w Br(L_w)$$is not a base change of a CSA over K. Proof. Assume that u is the base change of u'=\sum_v n_v v, i.e.$$u' \mapsto \sum_v \sum_{w|v} (\frac{4}{g_v} n_v )w =u$$Since for each of the$p_i$:$f=1$,$e=1$,$g=4$, we must have$n_{p_i}=\frac{1}{2^{k+2}}$. We know that$\sum_v n_v=0$(in$\mathbb{Q}/\mathbb{Z}$), and the contribution from the$p_i$'s is$2^k\frac{1}{2^{k+2}}=\frac{1}{4}$, hence at least one other place has its$n_v$'s denominator divisible by$4$. Since$Br(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$, this other place must be a finite place,$q$. If$q$is unramified, then by basic number theory$g_q\ne 1$($L$is a composite of two quadratic extensions,$L=K(\sqrt{a},\sqrt{b})$, so at least one of$a$,$b$,$ab$is a square modulo$q$). If$q$is ramified, then by the hypothesis,$g_q\ne 1$. This shows that$g_q \in \{2,4\}$. Hence, the denominator of$(\frac{4}{g_q})n_q$must be divisible by$2$, contradicting the fact that for any$w|q$,$n_w=0$in$u$. The smallest example, discriminant-wise (1521), of such an extension of$\mathbb{Q}$is the one cited above. (2) I won't pretend to understand anything about the last arrow. The only thing I will say is that the above proposition, proven with a bit of over-generalisation (if only the example is desired), gives a lot of CSA's that aren't base change. It is also interesting to note that if we remove the$g \ne 1$hypothesis, then I think that there is once again no obstruction, even though the$H^3\$ is not trivial.

3 removed nonsense
2 found smaller example
1