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(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(3) Some more general remarks on Q2: assume $K$ is a field and $A$ an abelian variety over $K$ such that $A(K)$ is finitely generated. Take any rational function $f$ on $A$. Then there is a subfield $K_0$ of $K$, finitely generated over the prime field, such that $f(A(K))\subset K_0$ (of course I exclude points where $f$ is undefined). Indeed, take for $K_0$ a field of definition for $A$, $f$, and generators of $A(K)$.

This proves that other examples for Q2 are: [EDIT: more generally, see (4) below]

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). Indeed, take $f$ defined and smooth at the origin, and (say) sending the origin to $0$. By the implicit function theorem, $f(A(K)$ contains a neighborhood $V$ of $0$ for the valuation topology; the subfield generated by $V$ is $K$.

(3b) Pseudo-algebraically closed (PAC) fields: indeed, by Bertini, there is an $f$ whose general fiber is smooth and geometrically connected, hence has rational points. (This works directly if $\dim A\geq2$; if $\dim A=1$, consider $A\times A$). This includes example (2) above.

(4) Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense. (Pop, who invented the concept, called these fields "large"; others say "ample").

I claim that every fertile field $K$ satisfies Q2. (This includes (3a) and (3b) above). Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$. Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$. It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blowing-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism $E\to\mathbb{P}^{g-1}_K$where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth. For every $y\in\mathbb{P}^{g-1}(K)$, $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$ also has rational points on $U\smallsetminus\{e\}$. Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective. On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).

[Edited November 2 for brevity]

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(3) A general result on Q2: Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense.
(Pop, who invented the concept, called these fields "large"; others say "ample").

I claim that every fertile field $K$ satisfies Q2. This includes in particular:

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial).

(3b) Pseudo-algebraically closed fields (i.e. such that every geometrically irreducible variety has a rational point). This includes example (2) above.

Proof of claim: Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$ (if $g=1$, consider $A\times A$). Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$. It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blow-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism $E\to\mathbb{P}^{g-1}_K$where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth.
For every $y\in\mathbb{P}^{g-1}(K)$, $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$ also has rational points on $U\smallsetminus\{e\}$. Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective.
On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(3) Some more general remarks on Q2: assume $K$ is a field and $A$ an abelian variety over $K$ such that $A(K)$ is finitely generated. Take any rational function $f$ on $A$. Then there is a subfield $K_0$ of $K$, finitely generated over the prime field, such that $f(A(K))\subset K_0$ (of course I exclude points where $f$ is undefined). Indeed, take for $K_0$ a field of definition for $A$, $f$, and generators of $A(K)$.

This proves that other examples for Q2 are:

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). Indeed, take $f$ defined and smooth at the origin, and (say) sending the origin to $0$. By the implicit function theorem, $f(A(K)$ contains a neighborhood $V$ of $0$ for the valuation topology; the subfield generated by $V$ is $K$.

(3b) Pseudo-algebraically closed (PAC) fields: indeed, by Bertini, there is an $f$ whose general fiber is smooth and geometrically connected, hence has rational points. (This works directly if $\dim A\geq2$; if $\dim A=1$, consider $A\times A$). This includes example (2) above.

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(3) Some more general remarks on Q2: assume $K$ is a field and $A$ an abelian variety over $K$ such that $A(K)$ is finitely generated. Take any rational function $f$ on $A$. Then there is a subfield $K_0$ of $K$, finitely generated over the prime field, such that $f(A(K))\subset K_0$ (of course I exclude points where $f$ is undefined). Indeed, take for $K_0$ a field of definition for $A$, $f$, and generators of $A(K)$.

This proves that other examples for Q2 are: [EDIT: more generally, see (4) below]

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). Indeed, take $f$ defined and smooth at the origin, and (say) sending the origin to $0$. By the implicit function theorem, $f(A(K)$ contains a neighborhood $V$ of $0$ for the valuation topology; the subfield generated by $V$ is $K$.

(3b) Pseudo-algebraically closed (PAC) fields: indeed, by Bertini, there is an $f$ whose general fiber is smooth and geometrically connected, hence has rational points. (This works directly if $\dim A\geq2$; if $\dim A=1$, consider $A\times A$). This includes example (2) above.

(4) Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense. (Pop, who invented the concept, called these fields "large"; others say "ample").

I claim that every fertile field $K$ satisfies Q2. (This includes (3a) and (3b) above). Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$. Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$. It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blowing-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism $E\to\mathbb{P}^{g-1}_K$where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth. For every $y\in\mathbb{P}^{g-1}(K)$, $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$ also has rational points on $U\smallsetminus\{e\}$. Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective. On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).

An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(3) Some more general remarks on Q2: assume $K$ is a field and $A$ an abelian variety over $K$ such that $A(K)$ is finitely generated. Take any rational function $f$ on $A$. Then there is a subfield $K_0$ of $K$, finitely generated over the prime field, such that $f(A(K))\subset K_0$ (of course I exclude points where $f$ is undefined). Indeed, take for $K_0$ a field of definition for $A$, $f$, and generators of $A(K)$.

This proves that other examples for Q2 are:

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). Indeed, take $f$ defined and smooth at the origin, and (say) sending the origin to $0$. By the implicit function theorem, $f(A(K)$ contains a neighborhood $V$ of $0$ for the valuation topology; the subfield generated by $V$ is $K$.

(3b) Pseudo-algebraically closed (PAC) fields: indeed, by Bertini, there is an $f$ whose general fiber is smooth and geometrically connected, hence has rational points. (This works directly if $\dim A\geq2$; if $\dim A=1$, consider $A\times A$). This includes example (2) above.