Why does $E\otimes_KE\cong EG$ imply that Galois theory works?

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 $E/K$. $E[G]$ is the group ring formed by finite group $G$.

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

(Solved by the comments below.)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? Does this extend to infinte galois extension case?

Ref: Algebra Vol 1: Fields and Galois Theory by Falko Lorenz

This question has been asked on stackexchange but there is no answer.(https://math.stackexchange.com/questions/2778581/why-e-otimes-ke-cong-eg-implies-that-galois-theory-works) Hence, I asked this on overflow.

$\textbf{Q':}$As qiaochu commented that this is related to torsor theory, I do not see why the two are related. It would be nice if someone could elaborate this a bit.

• If $E/K$ is finite then $|\text{Aut}(E/K)| \leq [E:K]$, with equality if and only if $E/K$ is a Galois extension. With this in mind, compute the $E$-dimension of both sides, letting $G$ be $\text{Aut}(E/K)$. – KConrad May 14 '18 at 18:44
• @KConrad I should deduce that $dim_E(E\otimes_KE)\geq dim_E(E[G])$ as $Aut(E/K)\leq[E:K]$. The content of the $E\otimes_KE\cong EG$ is equivalent to $E/K$ is galois for finite extension case. Is this correct? I also know $k[G]\cong E$ as $k[G]$ modules though non-canonical. So for finite extension, $k[G]\cong E$ iff $E[G]\cong E\otimes_KE$ iff $E/K$ galois. Is this interpretation correct? And what can I conclude about infinite extension via inverse limit of galois group? Can I pull out the inverse limit as direct limit on the other side? – user45765 May 14 '18 at 18:51
• Hendrik Lenstra has a result that generalizes the isomorphism $K[G]\cong E$ in the infinite-degree Galois case: think of the group ring as the functions $G \rightarrow K$. When $G$ is infinite, work with continuous functions ($G$ with the Krull topology and $K$ with the discrete topology). The title of his article is "The normal basis theorem for infinite extensions," or something close to that. – KConrad May 14 '18 at 18:59
• The assertion is equivalent to the normal basis theorem which some books do use to give the fundamental theorem of Galois theory. – Benjamin Steinberg May 14 '18 at 19:56
• For the relation to torsors in your second question, the point is that the normal basis theorem could be seen as a consequence of Galois descent for algebras. – F Zaldivar May 15 '18 at 14:52