A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$

Question

I think this is correct:

If $\phi:\mathbb Z^{n}\to\mathbb Z^{k}$ is a group homomorphism then $n=\operatorname{rank}\operatorname{im}\phi+\operatorname{rank}\ker\phi$.

Here is my attempt at a proof:

$\phi$ may be naturally and uniquely extended to a linear map between $\mathbb Q$-vector spaces $\widetilde{\phi}:\mathbb Q^{n}\to\mathbb Q^{k}$, where $n=\dim\operatorname{im}\widetilde{\phi}+\dim\ker\widetilde{\phi}$.

My questions:

Is this proof correct? I am concerned with "rank" either not making sense for subgroups or not coinciding with "dimension".
Does this appear in books (maybe with different formulation)? I have not found it in some standard textbooks I've looked in.

user26857 · Accepted Answer · 2015-01-08T20:54:01.737

1

No need to refer to $\mathbb Q$: we have the following short exact sequence of free $\mathbb Z$-modules (or abelian groups, if you like it) $$0\to\ker\phi\to\mathbb Z^n\to\operatorname{im}\phi\to0$$ which is split (helpful, but not necessary to conclude).

edited Jan 08 '15 at 20:54

answered Jan 08 '15 at 20:31

user26857

52,094

Is it correct that this short exact sequence is simply a formulation of the first isomorphism theorem? How does the rank equation follow from this short exact sequence? – Yoni Rozenshein Jan 08 '15 at 20:58
Yes. 2. If you know what means "split", then $\mathbb Z^n=\ker\phi\oplus\operatorname{im}\phi$ shows that $\mathbb Z^n$ has a basis resulting from the bases of $\ker\phi$ and $\operatorname{im}\phi$.

user26857

Jan 08 '15 at 21:01

I didn't know, but now I read about it. So as I understand: This sequence trivially left-splits; We use the splitting lemma to conclude $\mathbb Z^n = \ker \phi \oplus \operatorname{im} \phi$; and then we trivially get the union of bases as we wanted. So, the heart of the proof is in the proof of the splitting lemma. Right? – Yoni Rozenshein Jan 08 '15 at 21:08

@YoniRozenshein Yes. (This is because I don't want to refer to $\mathbb Q$, otherwise we can extend the maps to $\mathbb Q$-vector spaces with or without splitting.) – user26857 Jan 08 '15 at 21:11

Khako · Answer 2 · 2015-01-08T18:33:36.890

0

Since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, the extended $\mathbb{Q}$-linear map gives the split short exact sequence $$ 0\rightarrow \mathbb{Q}\otimes ker(\phi) \rightarrow \mathbb{Q}^n \rightarrow \mathbb{Q} \otimes Im(\phi) \rightarrow 0 $$ the result follows since $ker(\phi)$ and $Im(\phi)$ are free $\mathbb{Z}$-modules.

edited Jan 08 '15 at 18:33

answered Jan 08 '15 at 18:21

Khako

568

Recall that every non-zero submodule of a free module over a principle ideal domain is free as well. – Khako Jan 08 '15 at 18:24
I don't know the meanings of "flat", "split" and "$\otimes$" :( – Yoni Rozenshein Jan 08 '15 at 19:21
@YoniRozenshein :The notion "flat" is not needed: It is not hard to show that the extended $\mathbb{Q}$-linear map gives the short exact sequence $$ 0\rightarrow \mathbb{Q}\otimes ker(\phi) \rightarrow \mathbb{Q}^n \rightarrow \mathbb{Q} \otimes Im(\phi) \rightarrow 0. $$ This exact sequence is split since $\mathbb{Q}$ is a vector space. Then the result follows. – Khako Jan 09 '15 at 08:57

A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$

2 Answers2