Let $K / F$ be a finite separable field extension, and let $L / F$ be any field extension. Show that $K \otimes_{F} L$ is a product of fields.
My attempt: Since for any $f_1, f_2 \in F$ we have $f_1 \otimes f_2 = f_1f_2 (1 \otimes 1)$ it follows where an element of $K \otimes_F L$ sends the elements of $F$ is determined by where it sends $1 \otimes 1$.
Furthemore, we know that $K$ is seperable so that any element in $K$ has a minimal polynomial with distinct roots. By primitive element theorem, $K = F(\alpha)$ for some seperable $\alpha \in K$.
I'm not sure how to continue...
By the primitive element theorem, $K = F[x] / p(x)$ for some polynomial $p \in F[x]$. Then, $K \otimes_F L = L[x]/ p(x)$. Using that $L[x]$ is a UFD, factor $p(x)$ into a product of irreducible polynomials $\prod q_i^{e_i}$, with $q_i \in L[x]$ irreducible.
As $K$ is separable, $gcd(p, p') = 1$, so none of the $e_i$'s can be greater than 1. Finally use the Chinese remainder theorem.