Why is this isomorphism $M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$ an isomorphism of $M$ - algebras? Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $\textrm{Hom}_K(L,M)$ denote the set of all $K$ - algebra homomorphisms from $L$ to $M$. Since $L/K$ is separable we know that the number of elements in $\textrm{Hom}_K(L,M)$ is equal to $[L:K]$. Now I want to prove that we have an isomorphism of $M$ - algebras 
$$\varphi : M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$$
Proof that they are isomorphic as $K$ - modules : Write $L = K(a)$ for some $ a\in L$ (we can do this via the primitive element theorem). Then
$$\begin{eqnarray*} M \otimes_K L &=& M \otimes_K K[a] \\
&\cong& M\otimes_K K[x]/(f) \hspace{3mm} \text{where $f$ is the minimal polynomial of $a$ over $K$} \\
&\cong& M[x]/(f)\\
&\cong& M[x]/(f_1\ldots f_{[L:K]}) \hspace{3mm} \text{where the $f_i$ are the distinct}\\ && \hspace{1.5in} \text{irreducible factors of $f$ since $M/L$ is Galois} \\
&\cong& M^{[L:K]}\end{eqnarray*}$$
where  the last step was using the Chinese remainder theorem. For the third last step, we consider the ses 
$$ 0 \to (f) \to K[x] \to K[x]/(f) \to 0$$
and tensor with the exact functor $-\otimes_K M$ to get
$$0 \to (f) \otimes_K M \to K[x] \otimes_K M \to K[x]/(f) \otimes_K M \to 0$$
and so $$\begin{eqnarray*} M \otimes_K K[x]/(f) &\cong& K[x]/(f) \otimes_{K} M \\
&\cong& \frac{K[x] \otimes_{K} M}{f \otimes_K M}\\
& \cong& M[x]/(f) \end{eqnarray*}$$ where $(f)$ is now viewed as an ideal of $M[x]$.


My question is: The tensor product $M \otimes_K L$ is a left $M$ - module, but why is it also an $M$ - algebra? Also why are the isomorphisms above isomorphisms of $M$ - algebras and not just $K$ - modules?


 A: 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. 
