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?
 A: 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 $Br(K)$.
