# Why $E\otimes_KE\cong EG$ implies that Galois theory works?

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?

• It's a long story. The keyword you want to look up is "torsor." – Qiaochu Yuan May 13 '18 at 2:15
• @QiaochuYuan Is there a book that explains this story in detail? – Rene Schipperus May 13 '18 at 2:26
• Now posted to MO, mathoverflow.net/questions/300197/… – Gerry Myerson May 14 '18 at 23:06
• 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 $G$. – reuns Dec 18 '19 at 10:55

$$\def\Fix{\mathrm{Fix}}\def\Stab{\mathrm{Stab}}$$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. $$\square$$

• 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.) – darij grinberg Dec 19 '19 at 8:38
• 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.) – darij grinberg Jan 7 '20 at 15:00
• 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). – Captain Lama Jan 7 '20 at 15:51
• @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. – David E Speyer Jan 7 '20 at 16:04
• @darijgrinberg Thanks for the edits! I'll look at this in detail later. – David E Speyer Jan 7 '20 at 16:04