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[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$ has$q$). 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.
show/hide this revision's text 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.
show/hide this revision's text 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.
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