Vtyjkyuk

Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $textrmHom_K(L,M)$ 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 $textrmHom_K(L,M)$ is equal to $[L:K]$. Now I want to prove that we have an isomorphism of $M$ - algebras

$$varphi : M otimes_K L stackrelsimeqlongrightarrow M^[L:K]$$

Proof that they are isomorphic as $K$ - modules : Write $L = K(a)$ for some $ ain L$ (we can do this via the primitive element theorem). Then

$$begineqnarray* M otimes_K L &=& M otimes_K K[a] \
&cong& Motimes_K K[x]/(f) hspace3mm textwhere $f$ is the minimal polynomial of $a$ over $K$ \
&cong& M[x]/(f)\
&cong& M[x]/(f_1ldots f_[L:K]) hspace3mm textwhere the $f_i$ are the distinct\ && hspace1.5in textirreducible factors of $f$ since $M/L$ is Galois \
&cong& M^[L:K]endeqnarray*$$

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 $$begineqnarray* M otimes_K K[x]/(f) &cong& K[x]/(f) otimes_K M \
&cong& fracK[x] otimes_K Mf otimes_K M\
& cong& M[x]/(f) endeqnarray*$$ 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?

Let $S=Hom_K-alg(L,M)$ be the set of $K$-algebra morphisms $Lto M$ and consider the morphism of $K$-algebras $$f:Lto M^S:lmapsto (s(l))_s in S$$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 $$F:B=Motimes_K Lto M^S:b=motimes lmapsto (sigma_s(b))_sin S=(mcdot s (l))_sin S$$ Notice that the scalar multiplication by elements of $M$ on $Motimes_K L$ occurs via the left factor.

Notice also that we have used the canonical identification $$Hom_K-alg(L,M)stackrel =to Hom_L-alg(Motimes_KL,M):sstackrel =mapsto [sigma_s: motimes l mapsto ms(l)]$$ The morphism $F$ is surjective by the Reminder below.

On the other hand the domain $B=Motimes_K L$ of the morphism $F$ has dimension $operatorname dim_M Motimes_K L=[L:K]$ while its codomain $M^S$ has dimension $[M^S:M]=text cardS=[L:K]$ 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 $$F:Motimes_K Lto M^S:motimes lmapsto (mcdot sigma (l))_sigma in S$$ is an isomorphism of $M$-algebras.

Since $M^Scong M^[L:K]$ 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 $chi_j:Bto M$.

Then the resulting algebra morphism $F:Bto M^n:bmapsto (sigma_1(b),cdots, sigma_n(b))$ is surjective.

This results from the Chinese Theorem since the maximal ideals $operatorname Kersigma_j subset B$ are pairwise comaximal.