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.

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    $\begingroup$ I'm having a little difficulty parsing “It is perfect if the condition C is poseted for all $A \otimes_{\mathbb Z} k$.” The first layer of my confusion is whether you are referring to posets or if this is a typo for “posited”; presumably the latter, but I didn't want to make an assumption. $\endgroup$
    – LSpice
    Commented Jun 19 at 1:59
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    $\begingroup$ If $G=C_{p^\infty}$ then $G\otimes K=0$ for every field $K$. Hence the tensor products with fields will not make any difference between $C_{p^\infty}$ and $\{0\}$. $\endgroup$
    – YCor
    Commented Jun 19 at 9:07
  • $\begingroup$ @LSpice Sorry, this is a typo. I mean posted. $\endgroup$ Commented Jun 19 at 13:00

1 Answer 1


First note that it's not necessary to consider the tensor products $A \otimes K$ for arbitrary fields $K$; $K$ has some prime subfield $k$, and $A \otimes K \cong (A \otimes k) \otimes_k K$, so all the information is contained in the tensor products $A \otimes k$ for prime fields $k$, namely $\mathbb{Q}$ and $\mathbb{F}_p$ for primes $p$. So we can restrict our attention to these.

Second, there is no condition solely on the tensor products $A \otimes k$, $k$ prime, that is equivalent to $A$ being finitely generated. This is because we can find abelian groups $A, B$ such that $A \otimes k \cong B \otimes k$ for all prime fields $k$ but $A$ is finitely generated and $B$ isn't. Explicitly we can take $A = 0$ and $B = \mathbb{Q}/\mathbb{Z}$, both of which satisfy

$$A \otimes k \cong B \otimes k \cong 0$$

for all prime fields (and hence all fields) $k$. This condition on an abelian group means that it is torsion (which is equivalent to $A \otimes \mathbb{Q} = 0$) and divisible (which is equivalent to $pA = A$ for all primes $p$, which is equivalent to $A \otimes \mathbb{F}_p \cong A/pA \cong 0$).

  • $\begingroup$ Thanks Qiaochu. I think I also mention this kind of counter-example in my question. But what I want to know is a "condition" such that A\otimes K f.d. +the condition implies that A is finitely generated. $\endgroup$ Commented Jun 19 at 13:05
  • $\begingroup$ @Bingyu: I don't know what conditions are relevant to your situation because you haven't explained it. What kind of conditions can you impose if you can't impose "finitely generated"? Can you impose torsion-freeness, for example? $\endgroup$ Commented Jun 19 at 17:31
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    $\begingroup$ Also, I am sorry to be very pedantic, but the counterexample $\mathbb{Q}$ is different from the counterexample $\mathbb{Q}/\mathbb{Z}$. $A = \mathbb{Q}$ can be distinguished from a finitely generated abelian group because it satisfies $A \otimes \mathbb{Q} \cong \mathbb{Q}$ (which for a f.g. abelian group implies it has rank $1$) but $A \otimes \mathbb{F}_p \cong 0$ for all primes $p$ (which for a f.g. abelian group implies it has rank $0$). $\endgroup$ Commented Jun 19 at 17:33

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