Vector space with multiplication A vector space is a commutative group and I am wondering if it can be extended to be a ring by defining a multiplication. I tried $v \cdot w = (v_1 w_1, ..., v_nw_n)$ componentwise but then inverses aren't unique. Is it possible to construct a multiplication?
 A: The construction you described is a perfectly fine way to define a multiplication operation on a vector space, which is just viewing the vector space as the ring product of $n$ copies of the field.
There are other ways too, but they are not always possible for every $n$. Here's what I mean: suppose $n=k^2$. Then you can just rewrite the vectors as $k\times k$ matrices, and then you have multiplication given by matrix multiplication. That would be an entirely different multiplication than the one you proposed.
There are still more ways that a vector space can have a multiplication, but there are not a lot that work for generic vector spaces (the coordinatewise product above is an exception.) The study of vector spaces that are rings is just the study of algebras over fields.

I think there is another construction you should be interested in, but it is not exactly what you're asking for. This is purely for your information.
The tensor algebra $T(V)$ of a vector space $V$ is "the biggest algebra generated by $V$". In some sense this means it is a ring that contains $V$ and doesn't contain superfluous stuff, and that every other algebra like that is a quotient of it.
This produces a multiplication on a much larger set ($T(V)$) and when you multiply things in $V$ toegether, you do not generally get something back in $V$, so it is not really a multiplication on $V$. Still, this is a rather interesting thing to study. The tensor algebra itself, and several quotients of it (like the symmetric algebra, the exterior algebra, and Clifford algebras) are very interesting algebras.
A: What exactly do you mean by "inverses aren't unique. They are unique, if they exists. The multiplicative unit is $(1,1,\ldots, 1)$ and $v \in K^n$ has an inverse iff $v_i \ne 0$ for all $i$, the inverse is then given by $(v_1^{-1}, \ldots, v_n^{-1})$. The ring $K^n$ constructed is known as the direct product of $n$ copies of the ring $K$.
