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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?

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  • $\begingroup$ My question is: What motivates your question? The primitive element theorem is a basic tool in field theory. And your proof is standard. $\endgroup$ – Martin Brandenburg Dec 21 '12 at 12:35
  • $\begingroup$ @MartinBrandenburg Sorry I just realised I can do it without the primitive element theorem by inducting on the number of elements that we adjoint to $K$ to obtain $L$. I have changed my question now. I am sorry. $\endgroup$ – user38268 Dec 21 '12 at 12:36
  • $\begingroup$ Don't you naturally get an $L$-algebra structure on the tensor product of two $L$-algebras? The algebra being commutative, $M$ will inject into its center, so you get an $M$-algebra structure, no? $\endgroup$ – Jyrki Lahtonen Dec 21 '12 at 12:50
  • $\begingroup$ @JyrkiLahtonen You're saying that $M \otimes_K L$ is a $K$ - algebra (I get that) and hence an $M$ - algebra (which I don't get)? $\endgroup$ – user38268 Dec 21 '12 at 12:58
  • $\begingroup$ @JyrkiLahtonen how to use chinese remainder theorem in above proo given by user about modules? $\endgroup$ – maths student Aug 25 '18 at 15:26
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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.

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  • $\begingroup$ I always love it when you answer my questions because man that answer above was slick! I should have noticed that $M \otimes_K L$ was just a way of "changing our coefficients". However, that being said in my proof above is there a way to see why my sequence of isomorphisms are isomorphisms of $M$ - algebras? $\endgroup$ – user38268 Dec 21 '12 at 22:11
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    $\begingroup$ Dear BenjaLim, thanks for the kind words. My proof is, despite appearances, quite close to yours: the various $K$-homomorphisms $\sigma_i: L=K[a]\to M$ are obtained by sending $a$ to the various roots $a_i$ of the separable polynomial $f$ [in your notation $f_i=x-a_i$], so I think that you could prove that your isomorphism is one of $M$-algebras by following what happens to a decomposable tensor $m\otimes l \in M\otimes L$ along your displayed isomorphisms. $\endgroup$ – Georges Elencwajg Dec 21 '12 at 23:05
  • $\begingroup$ I like your proof in that it does not use the primitive element theorem. My proof will still work by induction (since any finite extension $L/K$ is always of the form $L = K(a_1,\ldots,a_k)$ for $a_1,\ldots,a_k \in L$) but of course an element-free approach is always nicest. I will get back to you if I have any problems with my isomorphisms above. $\endgroup$ – user38268 Dec 21 '12 at 23:26
  • $\begingroup$ I see the edit that you made above about Dedekind's theorem. Is this the same one on linear independence of characters as in page 51 of here? Thanks. $\endgroup$ – user38268 Dec 21 '12 at 23:57
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    $\begingroup$ Dear BenjaLim, Milne's independence of characters is a corollary of the more general version I quote, obtained by taking $L=K$, $A=K[G]$ (the group algebra) and using the adjunction formula $Hom_{grp}(G,K^*)=Hom_{K-alg}(K[G],K)$. $\endgroup$ – Georges Elencwajg Dec 22 '12 at 8:07

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