Tensor product of a vector space and a field

Question

Let $F$ be a field and $V$ a vector space of finite dimension $n$ over $F$. Let $\overline{F}$ be the algebraic closure of $F$. and let $\overline{V}=\overline{F}\otimes_F V$ the tensor product over $F$ of $V$ and $\overline{F}$.

In "A course of group theory", D.J.S. Robinson, p. 214, at the bottom of 8.1.10 it is said that "We can identify $a$ in $V$ with $1\otimes a$ in $\overline{V}$ ...".

I have some problems in doing that. In particular how can we prove that $1\otimes a$ is not a trivial element in the tensor product $\overline{V}$?

Answer 1

Every vector space over a field $K$ is the direct sum of copies of $K$. (possibly zero copies, or even an infinite cardinal number of them)

We can then use the fact the tensor product distributes over direct sums:

$$ \overline{F} \otimes_{F} \bigoplus_n F \cong \bigoplus_n \overline{F} \otimes_{F} F \cong \bigoplus_n \overline{F} $$

Answer 2

Since you're dealing with vector spaces (i.e. modules over a field) you have that for $V$ and $W$ two $F$-vector spaces with $\{v_1,v_2,\dots,v_\alpha\}$ and $\{w_1,w_2,\dots,w_\beta\}$ bases respectively of $V$ and $W$ (with $\alpha$ and $\beta$ ordinals), the family $\{v_i\otimes w_j \mid i = 1,\dots,\alpha, j =1,\dots,\beta\}$ is a basis for $V \otimes W$.

From this basic fact trivially follows that $1 \otimes a = 0$ if and only if $a=0$ in $V$, by putting $V=\bar F$ and $W=V$.