# Is a surjective linear operator on $\mathbb{Z}^n$ always bijective?

For the simple case of $n=1$, this is pretty easy since if $\phi\in Hom(\mathbb{Z})$ is surjective then $\exists z \in \mathbb{Z}$ such that $\phi(z)=1$, since $\phi$ is linear we can factor $\phi(z(1))=z\phi(1)=1$. Since $\phi$ must map $1$ to some integer, $z$ can only be $\pm1$ and thus $\phi$ is injective and therefore bijective. Can I generalize this argument?

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Hint: $\Z^n$ is a free $\Z$ module. – N. S. Apr 8 '12 at 20:18
A powerful generalization: a surjective endomorphism of a finite $A$-module is bijective. (Matsumura, Theorem 2.4) – Andrea Apr 10 '12 at 11:26

## 1 Answer

If $\phi\colon\mathbb Z^n\to \mathbb Z^n$ is surjective, then $\phi(e_1),\ldots,\phi(e_n)$ must span $\mathbb Z^n$. So the rank of the corresponding matrix must be $n$. Then, by the rank-nullity formula, it has to have zero kernel, meaning it is injective.

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I thought rank nullity doesn't hold over rings. – Steven-Owen Apr 8 '12 at 20:26
Think of $\mathbb Z$ as being embedded in $\mathbb Q$. If you are full rank over $\mathbb Z$, you are full rank over $\mathbb Q$. If you have trivial kernel over $\mathbb Q$, then you have trivial kernel over $\mathbb Z$. – Grumpy Parsnip Apr 8 '12 at 20:30
So the rank-nullity theorem should hold over any integral domain? – Fredrik Meyer Apr 8 '12 at 20:35
@FredrikMeyer: Looks like it to me. – Grumpy Parsnip Apr 8 '12 at 20:38
You should be careful to state what exactly the rank-nullity theorem over a domain should say since, over a general domain (not necessarily a PID), submodules of finite free modules need not be free. – KCd Apr 8 '12 at 21:08