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I am reading the book Algebra, volume 1: Fields and Galois theory by Falko Lorenz.

This is a part of statement in the book I do not fully appreciate. Suppose $E/K$ is Galois extension and $G$ the Galois group of $E/K$. Let $EG$ be the group algebra of the finite group $G$, considered as a $G$-module (not as a ring).

"It is worthwhile remarking that $E\otimes_KE\cong EG$ can be viewed as a deep reason why Galois theory works."

Q: What is the implication above? I though $E\otimes_KE\cong EG$'s proof has a major ingredient that the trace map is non degenerate.(i.e $E/K$ is separable.) Is this affording some representation of $G\to Aut_K(E)$? What is the author trying to express?

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    $\begingroup$ It's a long story. The keyword you want to look up is "torsor." $\endgroup$ – Qiaochu Yuan May 13 '18 at 2:15
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    $\begingroup$ @QiaochuYuan Is there a book that explains this story in detail? $\endgroup$ – Rene Schipperus May 13 '18 at 2:26
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    $\begingroup$ Now posted to MO, mathoverflow.net/questions/300197/… $\endgroup$ – Gerry Myerson May 14 '18 at 23:06
  • $\begingroup$ For $E/K$ Galois : the normal basis theorem is an isomorphism of $E[G]$-modules $E[G]=\sum_{\sigma\in G}[\sigma] E\to \sum_{\sigma\in G} \sigma(\alpha)K\otimes_K E = E\otimes_K E$, it is not an isomorphism of rings because $E[G]$ isn't commutative in general. In particular it means that the natural representation of $G$ on the $K$-vector space $E$ is the regular representation of $G$. $\endgroup$ – reuns Dec 18 '19 at 10:55
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$\def\Fix{\mathrm{Fix}}\def\Stab{\mathrm{Stab}}$I haven't read this book, but this statement makes sense to me. It only works, though, for a reader who thinks of certain facts about tensor products as standard. Here are the facts I am thinking of; all of them are easy to prove:

Let $L/K$ be an extension of fields and let $V$ be a $K$-vector space. Then:

Lemma 1: Let $G$ be a group acting $K$-linearly on $V$. Then $(L \otimes_K V)^G = L \otimes_K (V^G)$, where the superscript $G$ denotes the invariants.

Lemma 2: For any $K$-subspace $U \subseteq V$, we have $\dim_K U = \dim_L L \otimes_K U$.

Here is one more, which is not quite as easy:

Lemma 3: Let $X$ be a finite set and let $KX$ be the $K$-algebra of functions $X \to K$, with pointwise addition and multiplication. The unital $K$-subalgebras of $KX$ are in bijection with the equivalence relations on $X$; namely, an equivalence relation $\sim$ corresponds to the set of functions $f:X\to K$ which obey $f(x) = f(y)$ whenever $x \sim y$.

Proof Clearly, every equivalence relation gives a unital subalgebra and these subalgebras are all distinct. Conversely, let $A$ be a unital subalgebra and let $\sim$ be the equivalence relation on $X$ where $x \sim y$ if and only if, for all $f \in A$, we have $f(x) = f(y)$. We must show that $A$ contains all functions which obey $f(x) = f(y)$ whenever $x \sim y$.

It is enough to show that, for each equivalence class $C$, the subalgebra $A$ contains the function $e_C$ which is $1$ on $C$ and $0$ elsewhere. For each $y \not\in C$, there must be an $f \in A$ with $f(y) \neq f|_C$, and we can subtract off an element of $K$ and then rescale this $f$ so that $f(y)=0$ and $f|_C =1$. Multiplying together all such $f$'s, we get $e_C$. $\square$


Okay, now to answer the question. Let $E$ be a finite Galois extension of $K$, with Galois group $G$. This means, to us, that $E/K$ is a field extension of finite dimension, and $G$ is a finite group acting on $E$ by $K$-algebra automorphisms such that $E \otimes_K E \cong EG$ as $K$-algebras via the canonical homomorphism $E \otimes_K E \to EG$ that sends each $\alpha \otimes \beta \in E \otimes_K E$ to the function $G \to E, \ g \mapsto \alpha \cdot g(\beta)$. Note that the multiplication operation on the right is the pointwise multiplication of functions $G \to E$.

We note that, for $h \in G$, the action of $h$ on $EG$ takes $g \mapsto f(g)$ to $g \mapsto f(gh)$. We also let $G$ act on $E \otimes_K E$ by acting on the second tensorand (so $h\left(\alpha \otimes \beta\right) = \alpha \otimes \left(h\beta\right)$). The above isomorphism $E \otimes_K E \to EG$ is thus easily seen to be a $G$-module isomorphism.

We make $E \otimes_K E$ into an $E$-algebra by having $E$ act on the left tensorand (i.e., for $\alpha, \beta, \gamma \in E$, we set $\alpha\left(\beta\otimes\gamma\right) = \left(\alpha\beta\right)\otimes\gamma$). The above isomorphism $E \otimes_K E \to EG$ is thus an $E$-algebra isomorphism.

For a subgroup $H$ of $G$, let $\Fix(H)$ be the fixed subfield of $E$ (that is, the subfield $E^H$ of $E$); for a subfield $L$ of $E$, let $\Stab(L)$ be the stabilizer subgroup (i.e., the subgroup of $G$ consisting of all $g \in G$ such that $L \subseteq \Fix(g)$).

The first half of the Galois correspondence is the following:

Theorem 1 For any subgroup $H$ of $G$, we have $\Stab(\Fix(H)) = H$, and $\dim_K \Fix(H) = [G:H]$.

Proof By Lemmas 1 and 2, we have $\dim_K \Fix(H) = \dim_K E^H = \dim_E (E \otimes_K E)^H = \dim_E (EG)^H$, where the last equality sign follows from the $G$-module isomorphism $E \otimes_K E \to EG$. By our computation of the $G$-action on $EG$, the function $f : G \to E$ is fixed by $H$ if and only if $f(g) = f(gh)$ for all $h \in H$. In other words, $f$ must be constant on left $H$-cosets.

This makes it immediate that $\dim_E (EG)^H$ is $[G:H]$. Moreover, suppose that $h \in \Stab(\Fix(H))$. Then $h$ would also have to stabilize $E \otimes_K \Fix(H)$, which is $(E \otimes_K E)^H$ by Lemma 1, and thus we would have to have $f(gh)=f(g)$ for any function $f$ which is constant on left $H$-cosets. Clearly, this implies $h \in H$. $\square$

The other half of the Galois correspondence is

Theorem 2 Let $L$ be a subfield of $E$ containing $K$. Then $\Fix(\Stab(L)) = L$ and $[G:\Stab(L)] = \dim_K L$.

This one is a bit harder.

Proof Let $A = E \otimes_K L$. Then $A$ is clearly a unital $E$-subalgebra of $E \otimes_K E$, and thus (by Lemma 3) comes from an equivalence relation $\sim$ on $G$. Also, from $A = E \otimes_K L$, we have $\dim_E A = \dim_K L = \dim_K ((K \otimes_K E) \cap A)$ (since $(K \otimes_K E) \cap A = (K \otimes_K E) \cap (E \otimes_K L) = K \otimes_K L$ by elementary linear algebra).

Now, what is $K \otimes_K E$ in terms of the isomorphism $E \otimes_K E \cong EG$? A little thought shows that it is functions of the form $g \mapsto g(\beta)$, for $\beta \in E$ or, in other words, functions $f : G \to E$ obeying $f(g_1 g_2) = g_1 f(g_2)$.

Given a general equivalence relation $\sim$ on $G$ with corresponding $E$-subalgebra $A_{\sim}$ of $EG$, what is the intersection of $A_{\sim}$ with the space of such functions? Suppose that $g$ and $gh$ are equivalent under $\sim$ and consider any $g' \in G$. Then, any function $f$ in this intersection satisfies $f(g'h) = (g' g^{-1}) f(gh) = g' g^{-1} f(g) = f(g')$. Thus, if $H$ is the subgroup of $G$ generated by $g_1^{-1} g_2$ for all $g_1 \sim g_2$ and $f$ is a function as above, then $f$ must be constant on left $H$-cosets.

So, for a general equivalence relation $\sim$, we have $\dim_K A_{\sim} \cap (K \otimes_K E) = [G:H]$ in the above notation. On the other hand, $\dim_E A_{\sim} = |G/\!\sim\!|$.

For our particular $A$, we have $\dim_K A \cap (K \otimes_K E) = \dim_E A$. Therefore, $|G/\!\sim\!| = |G/H|$. The equivalence relation of being in the same $H$-coset coarsens $\sim$, so we have this equality if and only if $\sim$ is the equivalence relation by $H$-cosets. Thus, the stabilizer of $A$ is $H$, we have $(E \otimes_K E)^H = A$ and $\dim_E A = [G:H]$. Lemmas 1 and 2 allow us to transfer these back to the statements that the stabilizer of $L$ is $H$, that $E^H = L$ and that $\dim_K L = [G:H]$, as desired. $\square$

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  • $\begingroup$ How exactly do you define "Galois extension" and "Galois group" here? And is the $E \otimes_K E \to EG$ isomorphism supposed to be just a $K$-algebra homomorphism or also to be compatible with some kind of $G$-action? (Sorry, I'm being jetlagged and slow right now.) $\endgroup$ – darij grinberg Dec 19 '19 at 8:38
  • $\begingroup$ I've made a bunch of edits -- can you check I got your intent right? (It's somewhat tricky to figure out what exactly is being assumed, what follows and what holds anyway.) $\endgroup$ – darij grinberg Jan 7 '20 at 15:00
  • $\begingroup$ About Lemma 3, one should probably assume that $|K|\geqslant |X|$, otherwise $K$ does not have enough points for functions $X\to K$ to separate points in $X$, and we cannot recover all equivalence relations (in particular, the equality). $\endgroup$ – Captain Lama Jan 7 '20 at 15:51
  • $\begingroup$ @CaptainLama I think the proof is correct as written. Note my phrasing: "$x \sim y$ if and only if, for all $f \in A$, we have $f(x) = f(y)$". I do not require that there be one particular $f \in A$ which separates all the different equivalence classes from each other. $\endgroup$ – David E Speyer Jan 7 '20 at 16:04
  • $\begingroup$ @darijgrinberg Thanks for the edits! I'll look at this in detail later. $\endgroup$ – David E Speyer Jan 7 '20 at 16:04

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