This is an exercise given during my Commutative Algebra course. I reached to solve just the "if" arrow, but not the "only if". The question is:
Let $F\subseteq L$ be a finite degree extension of fields, say $[L:F]=n$, and consider the $L-$algebras $L\rightarrow L\otimes_{F}L$, given by $a\mapsto a\otimes 1$, and $L\rightarrow L^{n}$, given by $a\mapsto (a,\ldots,a)$. Then $L|F$ is a Galois extension if and only if $L\otimes_{F} L\simeq L^{n}$, where the isomorphism is an iso of $L-$algebras.
Now, I need to prove the "$\Leftarrow$" implication. I know the explicit form of the iso only on the first factor of tensor product because I know that the iso must commute with the $L-$algebra structure maps, hence, said $\psi$ the iso, we must have $\psi(a\otimes 1)=(a,\ldots,a)$. But I don't know how the iso work on a second factor different from $1$. I also know that being Galois means that the fixed field must coincide with $F$, or that the extension is normal and separable, or that $L$ is a splitting field of a separable polynomial with coefficients in $F$, or that the order of the Galois group $G$ of the extension must be $n$. I argued that any element $\sigma$ of Galois group must give rise to a commutative diagram of kind:
$$\begin{array} AL\otimes_{F}L & \stackrel{\psi}{\longrightarrow} & L^{n} \\ \uparrow{f} & & \uparrow{g} \\ L^{G} & \stackrel{\sigma}{\longrightarrow} & L^{G} \end{array} $$
where $f$ and $g$ are the restrictions of the structural map, but I don't know how to prove by this that $L^{G}=F$. Thanks for your suggestions.
It is a good exercise to compute $L \otimes_F F(a)$ first, where $a \in L$ is an element. The result is $\prod_i L[x]/(f_i^{v_i})$, where $f = \prod_i f_i^{v_i} \in F[x]$ is the minimal polynomial of $a$ and $f_i$ are its irreducible factors over $L$. By the way, this also proves directly $\Rightarrow$. And it shows that $a$ is separable iff $L \otimes_F F(a)$ is reduced (i.e. has no nilpotent elements except $0$).
Now assume that $L \otimes_F L \cong L^n$. Then this $L$-algebra is reduced. For every $a \in L$ the subalgebra $L \otimes_F F(a)$ is also reduced, hence $a$ is separable. This shows that $L/F$ is separable. By the primitive element theorem, we have $L=F(a)$ for some $a \in L$. Then $L^n \cong L \otimes_F L \cong \prod_i L[x]/(f_i)$ with the above notation. Now it is clear that the $f_i$ are linear, so that $f$ splits completely over $L$. This shows that $L/F$ is normal.