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Jason Starr
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By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of the map $$\text{Pic}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K})$$ equals the kernel of the pullback map on Brauer groups, $$\text{Br}(f):\text{Br}(\text{Spec}\ K) \to \text{Br}(C).$$ Of course $\text{Coker}(\text{Pic}(f))$ surjects onto the cokernel of the degree map, $$\text{deg}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K}/\text{Pic}^0_{C/K}) =\mathbb{Z}.$$ By the Snake Lemma, the kernel of the map $$\text{Coker}(\text{Pic}(f))\to \text{Coker}(\text{deg}(f))$$ equals the cokernel of your homomorphism, $$\text{Pic}^0(f):\text{Pic}^0(C)\to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}^0_{C/K}).$$ The cokernel of $\text{deg}(f)$ is cyclic. Thus, if $\text{Ker}(\text{Br}(f))$ is not cyclic, then $\text{Coker}(\text{Pic}^0(f))$ is nonzero. More precisely, for a prime $\ell$, if $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell \mathbb{Z}$ has rank $r+1$ as a vector space over $\mathbb{Z}/\ell\mathbb{Z},$ then $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$, possibly higher rank because of $\text{Tor}_1(\mathbb{Z}/\ell\mathbb{Z},\text{Coker}(\text{deg}(f))).$

The Brauer group of a global field is described by class field theory. In particular, for any prime $\ell,$ the rank of the $\ell$-torsion subgroup $\text{Br}(K)[\ell]$ is infinite (infinitude of primes). Let $\alpha_0,\dots,\alpha_r$ be $\mathbb{Z}/\ell\mathbb{Z}$-linearly independent classes in $\text{Br}(K)[\ell].$ Let $P_0,\dots,P_r$ be associated Severi-Brauer $K$-schemes. Now let $C$ be a general complete intersection curve in the product variety $$P:=P_0\times_{\text{Spec}\ K}\dots \times_{\text{Spec}\ K} P_r.$$ The kernel of the pullback map $\text{Br}(K)\to \text{Br}(P)$ contains the classes $\alpha_0,\dots,\alpha_r.$ Thus, the $\mathbb{Z}/\ell\mathbb{Z}$-vector space $\text{Coker}(\text{Pic}^0(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$.

By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of the map $$\text{Pic}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K})$$ equals the kernel of the pullback map on Brauer groups, $$\text{Br}(f):\text{Br}(\text{Spec}\ K) \to \text{Br}(C).$$ Of course $\text{Coker}(\text{Pic}(f))$ surjects onto the cokernel of the degree map, $$\text{deg}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K}/\text{Pic}^0_{C/K}) =\mathbb{Z}.$$ By the Snake Lemma, the kernel of the map $$\text{Coker}(\text{Pic}(f))\to \text{Coker}(\text{deg}(f))$$ equals the cokernel of your homomorphism, $$\text{Pic}^0(f):\text{Pic}^0(C)\to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}^0_{C/K}).$$ The cokernel of $\text{deg}(f)$ is cyclic. Thus, if $\text{Ker}(\text{Br}(f))$ is not cyclic, then $\text{Coker}(\text{Pic}^0(f))$ is nonzero. More precisely, for a prime $\ell$, if $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell \mathbb{Z}$ has rank $r+1$ as a vector space over $\mathbb{Z}/\ell\mathbb{Z},$ then $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$, possibly higher rank because of $\text{Tor}_1(\mathbb{Z}/\ell\mathbb{Z},\text{Coker}(\text{deg}(f))).$

The Brauer group of a global field is described by class field theory. In particular, for any prime $\ell,$ the rank of the $\ell$-torsion subgroup $\text{Br}(K)[\ell]$ is infinite. Let $\alpha_0,\dots,\alpha_r$ be $\mathbb{Z}/\ell\mathbb{Z}$-linearly independent classes in $\text{Br}(K)[\ell].$ Let $P_0,\dots,P_r$ be associated Severi-Brauer $K$-schemes. Now let $C$ be a general complete intersection curve in the product variety $$P:=P_0\times_{\text{Spec}\ K}\dots \times_{\text{Spec}\ K} P_r.$$ The kernel of the pullback map $\text{Br}(K)\to \text{Br}(P)$ contains the classes $\alpha_0,\dots,\alpha_r.$ Thus, the $\mathbb{Z}/\ell\mathbb{Z}$-vector space $\text{Coker}(\text{Pic}^0(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$.

By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of the map $$\text{Pic}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K})$$ equals the kernel of the pullback map on Brauer groups, $$\text{Br}(f):\text{Br}(\text{Spec}\ K) \to \text{Br}(C).$$ Of course $\text{Coker}(\text{Pic}(f))$ surjects onto the cokernel of the degree map, $$\text{deg}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K}/\text{Pic}^0_{C/K}) =\mathbb{Z}.$$ By the Snake Lemma, the kernel of the map $$\text{Coker}(\text{Pic}(f))\to \text{Coker}(\text{deg}(f))$$ equals the cokernel of your homomorphism, $$\text{Pic}^0(f):\text{Pic}^0(C)\to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}^0_{C/K}).$$ The cokernel of $\text{deg}(f)$ is cyclic. Thus, if $\text{Ker}(\text{Br}(f))$ is not cyclic, then $\text{Coker}(\text{Pic}^0(f))$ is nonzero. More precisely, for a prime $\ell$, if $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell \mathbb{Z}$ has rank $r+1$ as a vector space over $\mathbb{Z}/\ell\mathbb{Z},$ then $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$, possibly higher rank because of $\text{Tor}_1(\mathbb{Z}/\ell\mathbb{Z},\text{Coker}(\text{deg}(f))).$

The Brauer group of a global field is described by class field theory. In particular, for any prime $\ell,$ the rank of the $\ell$-torsion subgroup $\text{Br}(K)[\ell]$ is infinite (infinitude of primes). Let $\alpha_0,\dots,\alpha_r$ be $\mathbb{Z}/\ell\mathbb{Z}$-linearly independent classes in $\text{Br}(K)[\ell].$ Let $P_0,\dots,P_r$ be associated Severi-Brauer $K$-schemes. Now let $C$ be a general complete intersection curve in the product variety $$P:=P_0\times_{\text{Spec}\ K}\dots \times_{\text{Spec}\ K} P_r.$$ The kernel of the pullback map $\text{Br}(K)\to \text{Br}(P)$ contains the classes $\alpha_0,\dots,\alpha_r.$ Thus, the $\mathbb{Z}/\ell\mathbb{Z}$-vector space $\text{Coker}(\text{Pic}^0(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$.

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Jason Starr
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  • 111

By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of the map $$\text{Pic}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K})$$ equals the kernel of the pullback map on Brauer groups, $$\text{Br}(f):\text{Br}(\text{Spec}\ K) \to \text{Br}(C).$$ Of course $\text{Coker}(\text{Pic}(f))$ surjects onto the cokernel of the degree map, $$\text{deg}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K}/\text{Pic}^0_{C/K}) =\mathbb{Z}.$$ By the Snake Lemma, the kernel of the map $$\text{Coker}(\text{Pic}(f))\to \text{Coker}(\text{deg}(f))$$ equals the cokernel of your homomorphism, $$\text{Pic}^0(f):\text{Pic}^0(C)\to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}^0_{C/K}).$$ The cokernel of $\text{deg}(f)$ is cyclic. Thus, if $\text{Ker}(\text{Br}(f))$ is not cyclic, then $\text{Coker}(\text{Pic}^0(f))$ is nonzero. More precisely, for a prime $\ell$, if $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell \mathbb{Z}$ has rank $r+1$ as a vector space over $\mathbb{Z}/\ell\mathbb{Z},$ then $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$, possibly higher rank because of $\text{Tor}_1(\mathbb{Z}/\ell\mathbb{Z},\text{Coker}(\text{deg}(f))).$

The Brauer group of a global field is described by class field theory. In particular, for any prime $\ell,$ the rank of the $\ell$-torsion subgroup $\text{Br}(K)[\ell]$ is infinite. Let $\alpha_0,\dots,\alpha_r$ be $\mathbb{Z}/\ell\mathbb{Z}$-linearly independent classes in $\text{Br}(K)[\ell].$ Let $P_0,\dots,P_r$ be associated Severi-Brauer $K$-schemes. Now let $C$ be a general complete intersection curve in the product variety $$P:=P_0\times_{\text{Spec}\ K}\dots \times_{\text{Spec}\ K} P_r.$$ The kernel of the pullback map $\text{Br}(K)\to \text{Br}(P)$ contains the classes $\alpha_0,\dots,\alpha_r.$ Thus, the $\mathbb{Z}/\ell\mathbb{Z}$-vector space $\text{Coker}(\text{Pic}^0(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$.

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