# Can base-change be non-surjective on Brauer groups?

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?

## 1 Answer

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 $$Br(K)$$.

• 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. – RP_ Jan 17 '20 at 14:46
• 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. – GreginGre Jan 17 '20 at 18:09
• 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. – RP_ Jan 17 '20 at 20:34
• It is because quaternion algebras have exponent at most $2$, meaning $Q\otimes Q=0$ in the Brauer group. – GreginGre Jan 17 '20 at 22:56
• I have edited my answer and gave some details, in order to answer your question. – GreginGre Jan 18 '20 at 10:07