Let $K\subset L$ be a finite field extension of degree $n$.
Prove that it is Galois if and only if $L\otimes_{K} L\simeq L^{n}$, as $L$-algebras, considering on $L \otimes_{K}L=L^{n}$ the left structure of $L$-algebra $L \to L \otimes_{K} L $, $a \to a \otimes 1 $, and considering on the $n$-th fold product $L^{n}$ the structure of $L$-algebra $L \to L^{n}$, $a \to (a, \dots, a )$.
Any hint ?
Recall this question: $L$ is a separable $K$-extension if and only if $L \otimes_K L$ is reduced. So we just have to focus on normality.
By the primitive element theorem, $L = K [\alpha]$ for some $\alpha$ in $L$, where the minimal polynomial of $\alpha$ over $K$ is some $f (x) \in K [x]$ of degree $n$, and $f (x)$ splits over $L$ if and only if $L$ is a normal extension. But then $L \otimes_K L \cong L [x] / (f (x))$, so the Chinese remainder theorem implies $L \otimes_K L \cong L^n$ if and only if $f (x)$ splits over $L$. (The separability result mentioned above means we don't have to worry about repeated roots, so $f (x)$ will always factor as a product of pairwise coprime irreducibles.)