Why is this isomorphism $M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$ an isomorphism of $M$ - algebras?

Solution 1:

Let $S=Hom_{K-alg}(L,M)$ be the set of $K$-algebra morphisms $L\to M$ and consider the morphism of $K$-algebras $$f:L\to M^S:l\mapsto (s(l))_{s \in S}$$Since $M^S$ is an $M$-algebra, by the universal property of extension of scalars for algebras, $f$ extends to a morphism of $M$-algebras $$F:B=M\otimes_K L\to M^S:b=m\otimes l\mapsto (\sigma_s(b))_{s\in S}=(m\cdot s (l))_{s\in S}$$ Notice that the scalar multiplication by elements of $M$ on $M\otimes_K L$ occurs via the left factor.
Notice also that we have used the canonical identification $$Hom_{K-alg}(L,M)\stackrel {=}{\to} Hom_{L-alg}(M\otimes_KL,M):s\stackrel {=}{\mapsto} [\sigma_s: m\otimes l \mapsto ms(l)]$$ The morphism $F$ is surjective by the Reminder below.
On the other hand the domain $B=M\otimes_K L$ of the morphism $F$ has dimension $\operatorname {dim}_M M\otimes_K L=[L:K]$ while its codomain $M^S$ has dimension $[M^S:M]=\text {card}S=[L:K]$ too, because the extension $L/K$ is separable.
So $F$ is a surjective linear map between vector spaces of same $M$-dimension and is thus an isomorphism.
Conclusion
The morphism $$F:M\otimes_K L\to M^S:m\otimes l\mapsto (m\cdot \sigma (l))_{\sigma \in S}$$ is an isomorphism of $M$-algebras.
Since $M^S\cong M^{[L:K]}$ your question is answered, but in a more canonical way.

Reminder
Let $M$ be a field, $B$ an $M$-algebra and $\sigma_1,\cdots, \sigma_n$ a finite family of distinct $M$-algebra morphisms $\chi_j:B\to M$.
Then the resulting algebra morphism $F:B\to M^n:b\mapsto (\sigma_1(b),\cdots, \sigma_n(b))$ is surjective.
This results from the Chinese Theorem since the maximal ideals $\operatorname {Ker}\sigma_j \subset B$ are pairwise comaximal.