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I am reading Serre's book on Lie Algebras and Lie Groups. On Lemma 4.3, he states that

If $E$ is a finitely generated $\mathbb{Z}$-module and $\dim(E \otimes_{\mathbb{Z}} \mathbb{F}_p)$ over $\mathbb{F} = \mathbb{Z} / p\mathbb{Z}$ is independent of $p$, for all primes $p$, then $E$ is a $\mathbb{Z}$-free module with rank equal to the dimension of $E \otimes_{\mathbb{Z}} \mathbb{F}_p$ over $\mathbb{F}_p$.

Serra says that "this lemma is an easy consequence of the structure theorem of abelian groups". I have no idea on how to pass from $\mathbb{F}_p$ to $\mathbb{Z}$.

How can this be proved? Also, is there a nice "categorical" argument behind?

André Caldas
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  • But do you know the structure theorem of (finitely generated) abelian groups? Just write down a general such group with that theorem, and see what it must satisfy in order for that dimension equality to hold. – Torsten Schoeneberg Apr 10 '19 at 14:10
  • @TorstenSchoeneberg: Thank you for your reply! I think I see... $E = \mathbb{Z}^n \oplus \mathbb{F}p \oplus \dotsb$. Then $E \otimes\mathbb{Z} \mathbb{F}_p = \mathbb{F}_p^n \oplus \mathbb{F}_p \oplus \dotsb$. Is it right? – André Caldas Apr 10 '19 at 14:22

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The quoted theorem states that for a finitely generated $\Bbb Z$-module $M$, there are finitely many primes $p_1, ..., p_r$, as well as a number $k$, such that

$$ M \simeq \Bbb Z^k \oplus \bigoplus_{i=1}^r M_{p_i}$$

where $M_{p_i}$ is $p_i$-torsion.

Now notice that for two different primes $\ell \neq p$ and a $p$-torsion $M_p$, we have $\Bbb F_\ell \otimes M_p=0$, whereas $dim_{\Bbb F_p} \Bbb F_p \otimes M_p \ge 1$ iff $M_p \neq 0$.

Torsten Schoeneberg
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