TITLE: How to really understand the tensor algebra?
QUESTION [5 upvotes]: If $V$ is a vector space over $F$, then we define $T^r_0(V)=V^{\otimes r}$, then we define the algebra of contravariant tensors to be
$$T(V)=\bigoplus_{r=0}^\infty T^r_0(V)$$
together with the tensor product. But I'm really confused with that. The set $T(V)$ by construction, is the set of all sequences $T=(T_i)$ with $T_i=0$ for all but finitely many indexes $i\in \Bbb N$ with addition defined by $(T+S)_i = T_i+S_i$ and $(kT)_i=kT_i$.
The book I'm working with doesn't tell how the tensor product is defined in $T(V)$, but I think it is defined by $(T\otimes S)_i = T_i\otimes S_i$ which is the most natural thing to do. But I can't really understand how we should think about this space, it seems like to be putting together all kinds of contravariant tensors, but I can't really see why do we need this space, and how should we work with it.
I tried showing that this is really an algebra, but I'm confused if I did it right. I did the following: let $T,S\in T(V)$ and $a,b\in F$, then $(T\otimes (S+R))_i=T_i\otimes (S+R)_i,$ then from multilinearity of the tensor roduct on each factor $(T\otimes(S+R))_i=T_i\otimes S_i+T_i\otimes R_i$ wich means that $(T\otimes (S+R))_i = (T\otimes S)_i + (T\otimes R)_i$ and finally
$$(T\otimes (S+R))_i = ((T\otimes S)+(T\otimes R))_i\Longrightarrow T\otimes(S+R)=T\otimes S + T\otimes R,$$
the same for distribuitivity on the other side and for multiplication by scalars.
So, is this proof right? And in general, how should we really understand and think about the tensor algebra of a vector space?
Thanks very much in advance!

REPLY [3 votes]: There is a very simple underlying idea: we want to multiply vectors.
The algebra $T(V)$ gives us the most general way that we can multiply vectors while still retaining the rules of arithmetic, and contains nothing beyond what is needed to do so.
It's not hard to check that every element of $T(V)$ is, in fact, obtained by using scalars, vectors, addition, and multiplication. A little harder is to prove -- but very important to understand -- is the universal property that if $R$ is any ring over $F$, and $f : V \to R$ is an $F$-linear map, then there is a unique $F$-ring homomorphism $g : T(V) \to R$ that satisfies the property $g(v) = f(v)$ for all vectors $v$. (uniqueness is easy: it's the fact that it exists that is nontrivial)