TITLE: Is a vector space a ring, integral domain or field?
QUESTION [5 upvotes]: Is a vector space a ring, integral domain or field, with respect to  scalar multiplication? If you could give me an example, that would be awesome!

REPLY [3 votes]: The other way round, maybe: Every field (which is a ring and integral domain anyway) is a (one-dimensional) vector space over itself.
In general a vector space has no additional structure. You may view $\mathbb C$ as a two-dimensional vector space over $\mathbb R$ (or an infinite-dimensional vector space over $\mathbb Q$), bit the fact that $\mathbb C$ is also a field, has little if anything to do with that. 
And yet, some interesting vector spaces do carry an additional structure of multiplication so that this multiplication together with the addition the vector space has forms a ring. This mix of ring and vector space is an algebra. For example $\mathbb R[X]$, the polynomials in a variable $X$ with real coefficients, form a vector space over $\mathbb R$; but one can also multiply two polynomials to obtain a polynomial again, and this is compatible with polynomial addition. This makes $\mathbb R[X]$ an $\mathbb R$-algebra.
You could turn any vector space into an algebra by picking a basis and multiplying component wise . But thus depends on the choice of basis and hence is not natural.
With three dimensional space, we have something special: the cross product (and in contrast to what we have seen so far, this gives us a non-commutative ring and algebra!)