Question about the tensor algebra If $V$ is a vector space of finite dimension over $\mathbb{F}$, we define the tensor algebra: $T(V)= \oplus_{k=0}^\infty (\otimes^k V)$, where by convention $\otimes^0 V= \mathbb{F}$. My question is: why do we include $k=0$ in the direct sum? I see no reason for it, we would still have an algebra without it. The only possible reason I can see is if we include $k=0$, then we have a multiplicative identity, but this isn't really necessary as far as I can tell. 
I've been studying $T(V)$ in the context of differential geometry, so I see the necessity in including all powers $1 \leq m \leq n$, where $n= \dim(V)$. In order to keep the structure of an algebra, we would also have to keep all $k>n$, although we essentially get rid of this when we mod out by the ideal generated by elements of the form $v \otimes v$ to form the exterior algebra. But for $k=0$, I see no reason to keep it in the first place. Does anyone know of a specific reason for this?
 A: It is there so that this is an $F$-algebra; you need a ring homomorphism from $F$ to $T(V)$ in order to have this structure, and being an algebra over a field is a useful structural condition. You also just get better theorems if you don't mutilate $T(V)$ or $Sym(V)$ by removing the field!
But I guess you could clarify your question - necessary for what? 
From a geometric point of view, it is natural to consider functions as a simple case of tensor fields or differential forms. This is relevant if you are thinking about the various derivatives defined on these algebras, often these are defined on the degree 0 part in the usual way and then extended by some sort of Leibnitz rule. So the constant functions are just a natural part of this ring, and it simplifies things to keep them in it.
If you want to think about $Sym(V^*)$ as the algebraic functions on the affine space $V$, the constant functions are just there - they are the simplest functions! 
I can say more about modules over $k[t]$ than I can about modules over $k[t] \setminus k^*$, at least in part because the former are also vector spaces over $k$, and vector spaces are nice.
