# $\text{rank}(A')-\text{rank}(B')=\text{rank}(A)-\text{rank}(B)$ for free abelian groups.

I remember a proof I find about uniqueness of decompositions of finitely generated abelian group $A$ using the following lemma:

Let $p:A'\to A$ an epimorphism from free abelian groups of finite rank. Let $B$ be a subgroup of $A$ and $p^{-1}(B)=B'$. Then $$\text{rank}(A')-\text{rank}(B')=\text{rank}(A)-\text{rank}(B)\quad.$$

Unfortunately I cannot remember the proof, a bit difficult to find it alone, and neither the source.

It will be great if I someone knows a reference.

Edit: Ok I think I can prove that lemma, but I failed to see how it helps for uniqueness.

• Probably you mean $\DeclareMathOperator\rk{rank}\;\rk(A')-\rk(B')=\rk(A)-\rk(B)$? Oct 9, 2016 at 18:51
• @Bernard sorry I should ready more carefully before posting my question. Thanks Oct 9, 2016 at 18:52
• Personally, I should often read more carefully my answers before I post them :o) Oct 9, 2016 at 18:55

Hint:

Prove $A'/B'\simeq A/B$, then tensor with $\mathbf Q$, and use the fact that $$\DeclareMathOperator\rk{rank}\dim_{\mathbf Q}\bigl(A/B\otimes_{\mathbf Z}\mathbf Q\bigr)=\rk(A)-\rk(B)$$ and similarly for $A'/B'$.

Note:

$A/B\otimes_{\mathbf Z}\mathbf Q$ is (isomorphic to) the module of fractions $S^{-1}(A/B)$, with $S$ is the multiplicative subset of non-zero integers. In other words, it is the set of fractions $\dfrac{a+B}s$, where the numerator is a congruence class modulo $B$ and the denominator a non-zero integer. It becomes a $\mathbf Q$-vector space, with addition: $$\dfrac{a+B}s+\dfrac{a'+B}{s'}=\dfrac{s'a+sa'+B}{ss'}$$ and scalar multiplication: $$\dfrac pq\dfrac{a+B}s=\dfrac{pa+B}{qs}$$

• I don't what tensor is, the proof I saw was 'elementary'. Oct 9, 2016 at 19:29
• Do you know rings of fractions? I just use that abelian groups are $\mathbf Z$-modules, and we can extend scalar multiplication to $\mathbf Q$ so as to obtain a $\mathbf Q$-vector space. Oct 9, 2016 at 19:33
• Another variant, if you know the structure theorem for abelian groups: rank$(A)-$ rank$(B)$ is the free rank of $A/B$. Oct 9, 2016 at 19:36
• I've added some explanations but I'm not sure this be very clear. Oct 9, 2016 at 19:55
• your notation between $A/B$ and $\Bbb{Q}$ what does it means ? What do you mean by tensor with $\Bbb{Q}$ ? Oct 9, 2016 at 20:01