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?

Also, a simpler proof than >>20 gives:

Every field either has prime characteristic, or characteristic 0. Z cannot be a vector space over a field of characteristic p (p is prime), because that gives 0 = p*1z = (1f + 1f + ... + 1f)*1z = 1f*1z + 1f*1z + ... + 1f*1z = 1z + 1z + ... + 1z = p. (Here 1f denotes the multiplicative identity of the field, and 1z denotes 1 in the integers; 1f + 1f + ... + 1f has 1f occurring p times.) So we conclude that if Z is a vector space over a field F, F must have characteristic 0. However, if F has characteristic 0, let 2f = 1f + 1f. Since F is a field and 2f is necessarily non-zero, there exists a multiplicative inverse of 2f, we will call that (1/2)f. 2f * (1/2)f * 1z = 1z, and 2f * (1/2)f * 1z = 1f*(1/2)f*1z + 1f*(1/2)f*1z = (1/2)f*1z + (1/2)f*1z. So we have an element x = (1/2)f*1z of Z satisfying x + x = 1. This is, of course, not possible.