Let $L/K$ be a finite field extension. I need to prove that if $L/K$ is separable, then $E\otimes_KL$ is reduced for every algebraic field extension $E/K$.
I've readed that I need to use the Primitive Element Theorem and then the Chinesse Remainder Theorem, but I just don't see it. Any help would be appreciated.
Since $L/K$ is a finite separable extension, there is a $\theta\in L$ such that $L=K(\theta)$ by the primitive element theorem. Let $P$ be the minimal polynomial of $\theta$ over $K$. Let $E/K$ be any field extension, algebraic or not. Since $\theta$ is separable over $K$, the irreducible polynomial $P$ is a separable polynomial. This means that all the roots of $P$ in any extension of $K$ are simple. Let $$ P=P_1^{e_1}\cdots P_r^{e_r} $$ be the decomposition of $P$ in irreducible factors in $E[X]$. Since $P$ is separable, all $e_i$ are equal to $1$. It follows that $$ E\otimes_KL=E\otimes_K(K[X]/(P))=(E\otimes_KK[X])/(P)=E[X]/(P)=\prod_{i=1}^r E[X]/(P_i) $$ by the Chinese remainder theorem. Since the $P_i$ are irreducible in $E[X]$, the product above is a product of fields. In particular, it is reduced.