If $A$ is a finitely generated abelian group, then we know that for all fields $k$, the tensor product $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space.

The converse is not true, for example we can simply take $A=\mathbb{Q}$ as a counter-example.

So, is it possible to find conditions that make the converse could be true? i.e. I assume that $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space for all fields, and assume a condition C. Then $A$ is finitely generated.

It is perfect if the condition C is poseted for all $A\otimes_{\mathbb{Z}}k$. Since this is all the information I have. I will try other approaches to solve my original question (in my completely irrelevant research), but it is still interesting to know if there is any related content. It looks like an exercise, but I really didn't find any references.

If $A$ is a finitely generated abelian group, then we know that for all fields $k$, the tensor product $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space.

The converse is not true, for example we can simply take $A=\mathbb{Q}$ as a counter-example.

So, is it possible to find conditions that make the converse could be true? i.e. I assume that $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space for all fields, and assume a condition C. Then $A$ is finitely generated.

It is perfect if the condition C is posted for all $A\otimes_{\mathbb{Z}}k$. Since this is all the information I have. I will try other approaches to solve my original question (in my completely irrelevant research), but it is still interesting to know if there is any related content. It looks like an exercise, but I really didn't find any references.

If $A$ is a finitely generated abelian group, then we know that for all fields $k$, the tensor product $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space.

Conversely, this is not true, for example we can simply take $A=\mathbb{Q}$ as a counter-example.

So, is it possible to find conditions that make the converse could be true? i.e. I assume that $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space for all fields, and assume a condition C. Then $A$ is finitely generated.

It is perfect if the condition C is poseted for all $A\otimes_{\mathbb{Z}}k$. Since this is all the information I have. I will try other approaches to solve my original question (in my completely irrelevant research), but it is still interesting to know if there is any related content. It looks like an exercise, but I really didn't find any references.

If $A$ is a finitely generated abelian group, then we know that for all fields $k$, the tensor product $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space.

The converse is not true, for example we can simply take $A=\mathbb{Q}$ as a counter-example.

So, is it possible to find conditions that make the converse could be true? i.e. I assume that $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space for all fields, and assume a condition C. Then $A$ is finitely generated.

It is perfect if the condition C is poseted for all $A\otimes_{\mathbb{Z}}k$. Since this is all the information I have. I will try other approaches to solve my original question (in my completely irrelevant research), but it is still interesting to know if there is any related content. It looks like an exercise, but I really didn't find any references.