Is the set of integers a vector space?

My friend says it's not, because he thinks vectors have to have have at least two elements, otherwise they're scalars. I say that (Z, +, *) satisfies the axioms of a vector space and that's all you need.

What does /sci/ think?

>>16
Consider the vector space R^2 over the field R.  Does (1,0) have a multiplicative inverse? 

Here is an outline of a proof that Z is a not a vector space over any field. Suppose that there is a field F such that Z is a vector space over F.

Giving a vector space V over a field F is the same as giving a group V and a ring homomorphism F -> End(V), where End(V) is the ring of Endomorphisms of V.  In terms of Z, we need to determine End(Z).

Z is the free group on 1 generator, so any morphism Z -> Z is determined by the image of 1.  So it is not hard to show that End(Z) = Z as a ring.

This leads us to determining ring homomorphisms F -> Z.  Every homomorphism from a field is injective (F only has one ideal, (0)), so (F,+) is a subgroup of (Z,+).  But (Z,+) is a free group, so any subgroup is also free. Hence (F,+) = (Z,+) (note F is nontrivial), and the multiplication must also agree.  Hence F is isomorphic to Z, which is a contradiction, as F is a field.