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Is there a finite-degree separable field extension $\mathbb{K} \subset \mathbb{L}$ such that the induced map on Brauer groups $\operatorname{Br}(\mathbb{K}) \to \operatorname{Br}(\mathbb{L})$ is not a surjection?

I assume the answer is yes. What is an example?

Can it ever happen for finite fields? For number fields?

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For finite fields, the Brauer group is zero ( It comes from Wedderburn's theorem), so the answer is NO.

For number fields, the answer is YES. Following RP's question in the comments, I will prove the stronger statement that the map $Br(K)\to Br(L)^{Gal(L/K)}$ is not necessarily surjectve when $L/K$ is a Galois extension of number fields.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(i)$, and let $Q=(1+i,3)_L$. We have $\overline{Q}\simeq (1-i, 3)_L$, and thus $Q\otimes_L\overline{Q}\simeq (2,3)_L\simeq (-2,3)_L$, since $-1$ is a square in $L$. Now $3=1^2-(-2)1^2$ is a norm in $L(\sqrt{-2})$ so $(-2,3)_L$ is split. Therefore, $Q\otimes_L\overline{Q}\sim 0$, and since $Q\otimes_LQ\sim 0$ (it is a quaternion algebra), we get $Q\sim \overline{Q}$ , thus $Q\simeq \overline{Q}$ for degree reason.

Now it is a well-known fact that $Q$ is defined over $K$ if and only if $Cor_{L/K}(Q)\sim 0$ (it is a result specific to quadratic extensions and algebras of exponent 2: in fact, if $L/K=K(\sqrt{d})$, we have an exact sequence $H^1(K,\mu_2)\to H^2(K,\mu_2)\to H^2(L,\mu_2)\to H^2(K,\mu_2)$, where the maps are respectively cup-product by $(d),$ restriction, and corestriction.)

Now, we have $Cor_{L/K}(Q)\sim(N_{L/K}(1+i),3)_K=(2,3)_K$. Since $2$ is not a square mod $3$, the residue of $(2,3)_K$ at $3$ is non zero, hence $(2,3)_K$ is not split.

Consequently, the Brauer class of $Q$ lies in $Br(L)^{Gal(L/K)}$, but does not come from $Br(K)$.

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  • $\begingroup$ So a more subtle question could be whether there are examples such that $\operatorname{Br}(K) \to \operatorname{Br}(L)^\Gamma$ is non-surjective, where of course $\Gamma$ is the Galois group. $\endgroup$ – RP_ Jan 17 '20 at 14:46
  • $\begingroup$ I think there are counterexamples even for quadratic extensions. For exemple, if $L/K$ is quadratic with non trivial automorphism $*$ and $Q$ is a quaternion $L$-algebra, being in $Br(L)^\Gamma$ means $Q\otimes Q^*$ splits, that is $Res_{L/K}((Cor_{L/K})(Q))=0\in Br(L)$, or again that $ Cor_{L/K}(Q)$ is split by $L$, while being in the image of the restriction map means $Cor_{L/K}(Q)=0\in Br(K)$ (this is because $Q$ is a quaternion algebra, It does not work in general), that is $Cor_{L/K}(Q)$ splits over $K$. So the two things are different. $\endgroup$ – GreginGre Jan 17 '20 at 18:09
  • $\begingroup$ I am probably being dense, but why is the Galois invariance of the Brauer class the same as Q tensor Q* splitting? It's sort of counterintuitive to me, since by Hochschild-Serre I'd expect the cokernel of Br(K) -> Br(L)^\Gamma to live in some H^3, but your criterion would make it seem as if it was in some H^2 instead. But that's only using some general nonsense, you are probably using the specifics of the situation in some way I do not understand. $\endgroup$ – RP_ Jan 17 '20 at 20:34
  • $\begingroup$ It is because quaternion algebras have exponent at most $2$, meaning $Q\otimes Q=0$ in the Brauer group. $\endgroup$ – GreginGre Jan 17 '20 at 22:56
  • $\begingroup$ I have edited my answer and gave some details, in order to answer your question. $\endgroup$ – GreginGre Jan 18 '20 at 10:07

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