Tensor is non-zero Let $A,B$ be $k$-algebras, where $k$ is a field. Let $a_1,a_2,\dots a_n$ be elements of $A$, which are linearly independent over $k$. Similarly let $b_1,b_2, \dots, b_n$ be elements of $B$, which are linearly independent over $k$. Is $\sum a_i \otimes_k b_i$ non-zero in $A \otimes_k B$?

On first look the claim seems to be true. As far as I know $\sum a_i \otimes_k b_i =0$ if and only if we can bring $ \sum a_i \otimes_k b_i$ to the $0 \otimes_k 0$ tensor using the tensor-defining relations: 
$$(a+a',b) = (a,b) + (a',b); \quad \quad (a,b+b') = (a,b)+(a,b'); \quad \quad (ka,b) = (a,kb);$$
However, I feel the linear independence really prohibits us from bringing the sum to a simple tensor along either of the coordinates, unless $n=1$. But, I'm struggling to prove that this is indeed the case if $n\ge2$. While if $n=1$, $a \otimes_k b = 0$ implies that $a=0$ or $b = 0$, as $A$, $B$ are flat, given that $k$ is a field. Obviously this contadicts the linear independence.
Also I tried to use the flatness of $A$ and $B$, however to no avail.
Finally, if it makes the problem simplier we may assume that $A$ is an integral domain and $B$ is a finite field extension of $k$ with $\{b_1, \dots, b_n\}$ a basis of $B$ over $k$. In fact, this is the problem I initially wanted to solve, but I thought that the result might be somewhat generalized.
 A: Let's back up to a special case where $A, B$ are vector spaces, with respective bases $\{a_j\}, \{b_j\}$.
Furthermore, let's define the tensor product $A \otimes B$ (I'll omit the subscript $k$ for brevity) as the dual of the space ${\tt Bil}(A, B)$ of all bilinear forms on the direct sum $A \oplus B$.  Namely, $a \otimes b$ is the linear functional mapping each bilinear form $\psi( \cdot, \cdot)$ on $A \oplus B$ to the scalar $\psi(a, b)$.  [I find this, geometric, definition a lot friendlier to work with.]
How does $\sum_{i}a_{i} \otimes b_{i}$ act on a bilinear form $\psi( \cdot, \cdot) \in {\tt Bil}(A, B)$? By mapping that form to
$$
\sum_{i} \psi( a_{i}, b_{i} ).
$$
Can we construct a $\psi$ so that the latter sum is not zero?  If so, then the linear functional $\sum_{i}a_{i} \otimes b_{i}$ on ${\tt Bil}(A, B)$ cannot be identically zero.
A: We even do get somewhat more! If you have $ k $ vector spaces
$ V, W $
with bases
$ (v_i)_{i \in I} $
and
$ (w_j)_{j \in J} $,
then 
$ (v_i \otimes w_j)_{i,j} $
is a basis of 
$ V \otimes W $.
Now this implies the statement, because in the context of your question 
$ a_1, \dots a_n $ and $ b_1, \dots, b_n $
lie in a basis of 
$ A $
and 
$ B $
respectively. Therefore by the upper statement
$ (a_i \otimes b_j)_{i,j} $
is a subfamily of a basis of 
$ A \otimes B $
and therefore in particular linearly independent. Thus 
$ \sum_i a_i \otimes b_i \not = 0 $.
Here are some detailed hints for the first statement. But if you want to, give it a try yourself. Probably the most accessible setting is to try to construct an isomorphism 
$ \oplus_I k \times \oplus_J k \to \oplus_{I \times J}k $
through the universal property of the tensor product.
Our aim is to construct an isomorphism
$ V \otimes W \cong \oplus_{I \times J}k $.
Through the choice of bases we have an isomorphism
$ V \times W \cong \oplus_I k \times \oplus_J k $.
Now we can define the map
$$ \oplus_I k \times \oplus_J k \to \oplus_{I \times J} k, (\sum_i \lambda_i \delta_i, \sum_j \mu_j \delta_j) \mapsto \sum_{i,j} \lambda_i \mu_j \delta_{i,j}.  $$
whereby the
$ \delta $
-s denote the standard basis vectors. It is not too hard to see, that this map is bilinear. By the universal property of the tensor product this yields a unique liner map
$ V \otimes W \to \oplus_{I \times J}k $.
To show, that this is truly an isomorphism we need an inverse map. Remember that we want, the family
$ (a_i \otimes b_j)_{i,j} $
to be a basis of
$ V \otimes W $.
So what could be more natural than to try the map given by
$$ \oplus_{I \times J}k \to V \otimes W, \delta_{i,j} \mapsto v_i \otimes w_j? $$
To convince yourself, that this defines an inverse should be another good exercise.
A: Let $k$ be a field, and $V$ and $W$ two $k$-vector spaces, with bases $e_i$ and $f_j$ respectively. Then $V\otimes_kW$ is a $k$-vector space with basis $e_i\otimes f_j$. Thus the answer to your question is yes.
One way to see this result about bases is to use the universal property. A $k$-bilinear map $V\times W\to X$ is completely determined by its values on $(e_i,f_j)$, and we can choose these values freely.
