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?


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 $Br(K)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.