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The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

Motivation

###Motivation TheThe canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

###Motivation The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

Motivation

The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

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James D. Taylor
James D. Taylor
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The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an infinitely generated algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

###Motivation The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, it$\mathbb{R}$ is not an infinitely generated algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an infinitely generated algebraic extension of $k$. Does this imply that $X$ has infinitely many $K$-points?

###Motivation The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, it is not an infinitely generated algebraic extension of a field for which this curve does have a rational point.

The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

###Motivation The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

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James D. Taylor
James D. Taylor
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Does a curve have infinitely many $K$-rational points under these hypotheses?

The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an infinitely generated algebraic extension of $k$. Does this imply that $X$ has infinitely many $K$-points?

###Motivation The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, it is not an infinitely generated algebraic extension of a field for which this curve does have a rational point.