Let $F|K$ be a finite field extension and let $E|K$ be an algebraic extension. I want to prove that if $\{a_{1},\ldots,a_{n}\}$ is a $K$-basis of $F$, then $\{1\otimes a_{1},\ldots,1\otimes a_{n}\}$ is an $E$-basis of $E\otimes_{K} F$.
We are going to see that it is a generating set. It is enough if we prove that the elements of the form $e\otimes f\in E\otimes_{K}F$ are generated by that set. Let $k_{1},\ldots,k_{n}\in K$ such that $$ f=\sum_{i=1}^{n}k_{i}a_{i}\Rightarrow e\otimes f=\sum_{i=1}^{n}ek_{i}(1\otimes a_{i}). $$ But how do we show that $\{1\otimes a_{1},\ldots,1\otimes a_{n}\}$ are $E$-linearly independent?
I am fairly certain this proof works. Typically, a good way to get a hold of the tensor product is by looking at its dual space. Let $\{ a_1^*,...,a_n^* \}$ be the dual basis for $F$ as a $k$-vector space, and consider the functions $$1 \otimes a_{i}^*: E \otimes_k F \rightarrow E \otimes_k k \cong E.$$ These will be $k$-linear maps, but will also be $E$-linear maps, as $$ (1 \otimes a_{i}^*)(\lambda (e \otimes f)) = (\lambda e) \otimes a_i^*f = \lambda (e \otimes a_{i}^*f) = \lambda (1 \otimes a_{i}^*) (e \otimes f). $$ Also, these maps distinguish each $1 \otimes a_j$, as $(1 \otimes a_i^*)(1 \otimes a_{j}) = 1 \otimes \delta_{ij} = \delta_{ij} \otimes 1,$ where $\delta_{ij}$ is the Kronecker-delta function. The isomorphism from $E \otimes_k k$ to $E$ maps $e \otimes 1$ to $e$, so this will be zero in $E$ precisely if $i = j$.
Now, if the set were $E$-linearly dependent, then we would have $\sum_{i} e_i (1 \otimes a_i) = 0.$ But applying each of these $E$-linear-functionals, we find that each $e_i =0$.