Can we find all group homomorphism from $\mathbb{Z}^n$ to $\mathbb{Z}^n$?
For such a map surjective always imply isomorphism (like $\mathbb{Z}$)?
Asked
Active
Viewed 357 times
2
-
do you mean the free abelian product? – Daniel Valenzuela Aug 07 '14 at 10:53
-
usual direct sum of n copy of integers – jeevan Aug 07 '14 at 10:54
-
Look at the proof of how linear maps of vector spaces (modules over a field) are in one-to-one correspondence to matrices (with fixed basis). The same will work here. – Daniel Valenzuela Aug 07 '14 at 10:56
-
what answer should I expect? – jeevan Aug 07 '14 at 11:08
-
you know how all linear maps look like – Daniel Valenzuela Aug 07 '14 at 11:10
-
okay, thanks... – jeevan Aug 07 '14 at 11:25
-
Do you know what an endomorphism ring is? And do you know that the automorphism group is $\operatorname{GL}_2(\mathbb{Z})$? – user1729 Aug 07 '14 at 12:19
-
This question is covered by the one suggested as duplicated. However, the present question can be encountered (and handled) much earlier in the course of one's studies, so I decided to vote "leave open". – Jyrki Lahtonen Aug 12 '14 at 07:48
2 Answers
2
if the epic morphism were not monic then it would have a non-trivial kernel $K$. then, we would have $Z^n \cong Z^n \oplus K$
David Holden
- 18,040
-
-
2you need to tell us what you know. The statement is clear when you know about split exact sequences. Otherwise you may know about the isomorphism theorem, so that a surjective map $Z^n \to Z^n$ induces an isomorphism $Z^n/K \to Z^n$ where $K$ is the kernel. Since the rank is an invariant (preserved by isomorhisms) as soon as the kernel $K$ is non trivial, the map would not be an isomorphism. Therefore surjectiveness implies trivial kernel; hence an epimorhism is automatically an isomorphism. – Daniel Valenzuela Aug 07 '14 at 11:08
-
thx Dan. i am not on very sure ground as yet! my thought was that $Z^n$ is a projective $Z$-module and that would ensure that an exact sequence splits. however i am not very clear on what "projective" means. from the little i have read it seems to be to do with having a basis. your comment suggests it is the finite rank which is the crux of the matter. could you suggest a good short ref which might help me to sharpen up my understanding? – David Holden Aug 07 '14 at 11:16
2
$\operatorname{Hom}_{\mathbb Z}(\mathbb Z^n,\mathbb Z^n)\simeq\operatorname{Hom}_{\mathbb Z}(\mathbb Z,\mathbb Z^n)^n\simeq\mathbb Z^{n^2}$. In fact, a group homomorphism from $\mathbb Z^n$ to $\mathbb Z^n$ corresponds to a matrix in $M_{n}(\mathbb Z)$.
Every surjective endomorphism of $\mathbb Z^n$ is an isomorphism: if $A\in M_n(\mathbb Z)$ is the corresponding matrix of a surjective endomorphism of $\mathbb Z^n$, then there exists $B\in M_n(\mathbb Z)$ such that $AB=I_n$. Now, by taking the determinant, it follows that $A$ is invertible. (For a more general case see here.)
-
thx @user26857, that link is a great introduction to Nakayama's Lemma – David Holden Aug 17 '14 at 08:48