abstract algebra - 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 $A \otimes_k B$?

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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.

Answer

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.