Return to Answer

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 1$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the higher Galois cohomology of a finite field with coefficients in $\mathbf{G}_m$ is zero (by an argument similar to the Lemma above using the Kummer sequence), so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 1$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the higher Galois cohomology of a finite field with coefficients in $\mathbf{G}_m$ is zero (by an argument similar to the Lemma above using the Kummer sequence), so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{0,2}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 0$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the higher Galois cohomology of a finite field with coefficients in $\mathbf{G}_m$ is zero (by an argument similar to the Lemma above using the Kummer sequence), so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 1$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the higher Galois cohomology of a finite field with coefficients in $\mathbf{G}_m$ is zero (by an argument similar to the Lemma above using the Kummer sequence), so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 0$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the Galois cohomology of a finite field in odd degree is zero, so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.

This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global function fields.

Theorem: Define $k := \mathbf{F}_q$ and let $X/k$ be a smooth proper geometrically connected curve with function field $K$. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$ The sum in the middle is over all (non-archimedean) places of $K$.

Let us first prove the following.

Lemma: Let $A/k$ be an abelian variety. Then for all $i \geq 1$, $$H^i_{\text{ét}}(k,A) = 0.$$ When $i = 1$, this is a special case of Lang's theorem. Now for $i > 1$, let $n \in \mathbf{N}$ be any integer, and consider the exact sequence in the fppf topology $$ 0 \to A[n] \to A \stackrel{n\cdot }{\to} A \to 0.$$ Now I claim that $H^i_{\text{fppf}}(k, A[n]) = 0$ for all $i > 0$. Indeed, since $k$ is perfect, we have (see here) $$H^i_{\text{fppf}}(k, A[n]) = H^i_{\text{ét}}(k, A[n])$$ which vanishes for cohomological dimension reasons. It follows that multiplication by $n$ on $H^i_{\text{ét}}(k, A)$ is an isomorphism. On the other hand, $H^i_{\text{ét}}(k,A)$ is also torsion and therefore must be zero since $n$ was arbitrary.

We can now prove the theorem. From now on all cohomology considered will be étale cohomology. Consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0$$ where $\eta$ is the generic point of $X$, $j: \eta \to X$ the canonical inclusion, and the sum on the right is over all closed points of $X$. Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$

We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& \mathbf{Q}/\mathbf{Z}\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$

I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} k$$ denote the structure map. Consider the Leray spectral sequence with second page $E_2^{i,j} := H^i(k, R^jf_\ast \mathbf{G}_m)$ converging to $E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$ We are interested in computing the $E_\infty^2$ term, namely $H^2(X,\mathbf{G}_m)$. By construction of the Leray spectral sequence, this admits a filtration $$0 = F^{-1} \subseteq F^0 \subseteq F^1 \subseteq F^2 = E_\infty^2$$ with $$\begin{eqnarray} F^0/F^{-1} &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ F^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(k, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the (Azumaya) Brauer group of $k$, which we know is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and thus $F^0/F^{-1} = 0$.

Now we show that $F^1/F^0 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the higher Galois cohomology of a finite field with coefficients in $\mathbf{G}_m$ is zero (by an argument similar to the Lemma above using the Kummer sequence), so it is sufficient to show that $E_2^{1,1} = H^1(k, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/k$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/k} \to \text{Pic}_{X/k} \to \mathbf{Z} \to 0,$$ where the last map is $\mathcal{L} \mapsto \text{deg}(\mathcal{L})$. Taking cohomology, we have an exact sequence $$H^1(k, \text{Pic}^0_{X/k}) \to H^1(k, \text{Pic}_{X/k}) \to H^1(k, \mathbf{Z}).$$ The left term is zero by the lemma above. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{k}/k)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.

Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{k} \to \operatorname{Spec}k$ is the Brauer group of $\overline{X} := X \times_{k} \overline{k}$. But $\overline{X}$ is a regular integral scheme, and hence $\text{Br}(\overline{X})$ must inject into the Brauer group of its function field which is zero by Tsen's theorem.