Q:

$0^{th}$ tensor power, $V^{\otimes 0} = \Bbb F$, definition, or mathematical construction?

95% sure I will be told it's just a definition, move on etc:

Is there mathematical reason why the $0^{th}$ tensor power is defined as:
$$V^{\otimes 0} = \Bbb F$$

Where $\Bbb F$ is the field that vectorspace $V$ lies over.

We define the $n^{th}$ tensor power of $V$ as:
$$ V^{\otimes n}=V\otimes V\otimes \cdots \otimes V$$
with basis:
$$\{v^{i_1}\otimes v^{i_2} \otimes \cdots \otimes v^{i_n}: 1\leq i_k \leq n\}$$

For some reason I feel like $V^{\otimes 0}$ should equal the trivial vectorspace.

So is it just a definition? If so, what is the motivation for this choice. Thanks.

A:

You want it to be true that $V^{\otimes(n+m)} \cong V^{\otimes n} \otimes V^{\otimes m}$, which means that $V^{\otimes 0}$ has to be the unit for the tensor product, which is the underlying field. This is a straightforward categorification of $a^0 = 1$. 
Alternatively, you can think of $V^{\otimes 0}$ as having basis the "empty tensor product" of zero vectors in $V$. It helps to think of this as the unit in the tensor algebra $T(V) = 1 \oplus V \oplus V^{\otimes 2} \oplus \dots $. 

