Q: Natural extension of basis for a field extension is a basis for tensor product Suppose $A$ is a $k$-algebra, and $l/k$ a finite field extension. 

Claim: if $b_1 =1 , b_2,\ldots, b_n$ is a $k$-basis for $l/k$, then $1 \otimes b_1, \ldots, 1 \otimes b_n$ is a free $A$-basis for $A \otimes_k l$.

It is clear that these are a generating set with coefficients in $A$, but why are they $A$-independent?

A: I was just being dumb, thanks reuns. Here is the solution for closure.
Take a basis $A = \sum_j ka_j$ (not necessarily finite). Then the $a_j \otimes_k b_m$ are a $k$-basis for the tensor product. Assume we have
$$
   0 = \sum_{i} a_i' (1 \otimes b_i)  
$$
for arbitrary $a_i' \in A$. Then
$$
0 = \sum_{i,j} c_{i,j} (a_j \otimes b_i)
$$ 
with only finitely many $c_{i,j} \neq 0$. Then we have all $c_{i,j} = 0$, so $a_i' = \sum_j c_{i,j} a_j = 0$ for all $i$.
