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I need a reference which states which of the "normal properties of vector spaces" carry over to free $\mathbb{Z}$-modules.

Especially I am interested in things like: If you have a linear map between two free $\mathbb{Z}$-modules and you choose a basis for its kernel, can you choose a basis of a complementary space so that both together form a basis of the whole space (and the map, viewed only on this complementary space, is an isomorphism on its image)?

Probably this is an easy question for algebra guys.

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To make my question more precise: What are important properties of subspaces of Z^n, regarding bases, complements, kernels and cokernels? Is there a good book about that? – J. Fabian Meier May 12 '10 at 11:33
I first learnt the basics of this from the textbook… which now alas is out of print. – Robin Chapman May 12 '10 at 12:04
This comment is just to emphasize that Smith normal form (as in bugs' answer below) really is a good way to answer the types of questions you ask about (for any particular map between finite rank modules). The wikipedia entry is, I think, pretty readable: – GS May 12 '10 at 18:49
up vote 3 down vote accepted

What carries over?

As Peter pointed out, a submodule of a free $\mathbb{Z}$-module though free need not have a complement. Indeed each submodule of a free $\mathbb{Z}$-module is free, but a quotient module need not be, for instance $\mathbb{Z}/2\mathbb{Z}$. Also a $\mathbb{Z}$-module is free if and oly if it is projective; this entails that a kernel of a map of free modules does have a complement.

The set $\mathrm{Hom}(F,G)$ for free $\mathbb{Z}$-modules need not be free. If $F$ is free of countably infinite rank and $G=\mathbb{Z}$, then $\mathrm{Hom}(F,G)\cong\prod_{j=1}^\infty\mathbb{Z}$ which remarkably is not free over $\mathbb{Z}$. But $F\otimes G$ is free for free $F$ and $G$.

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So what is an example of a map f:Z^n -> Z^m, where ker f has no complement V, so that f|V is an isomorphism onto the image of f? – J. Fabian Meier May 12 '10 at 11:48
The kernel of a map between free Z-modules always has a complement (the image is a submodule of a free module, hence itself free, choose a splitting). – user2035 May 12 '10 at 11:57
Now corrected – Robin Chapman May 12 '10 at 11:58

You can write your map as a matrix. Moreover, you can choose a different basis so that the matrix is in the Smith's normal formal. The complement to the kernel exists only if the Smith's normal form contains only ones and zeroes. That is about it and should be explained in many Algebra books, say, Artin's Algebra.

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I think you meant to say that the complement to the image exists iff the SNF has only zeroes and ones. – Robin Chapman May 12 '10 at 12:03
Yeah, good point! My statement is still correct but the next statement "that is about it" would not be correct without iff:-)) – Bugs Bunny May 12 '10 at 12:13
Bugs, why do you say "That is about it" rather than "That's all, Folks"? It almost makes me think you might not be the real Bugs Bunny. – KConrad May 12 '10 at 22:28
Of course you realize, this means war. – Bugs Bunny May 21 '10 at 9:11

There is no complementary space in general: Consider the multiplication by 2 map from $\mathbb{Z}$ to itself ...

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I did not claim that. I was just interested in complementary spaces to kernels of linear maps. If we do not exist in general, I don't care: I was searching for a reference which states what kind of vector space (basis related) theorems also work (in a maybe weaker sense) for free Z-modules. – J. Fabian Meier May 12 '10 at 11:30

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