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Edited (in view of new comments to this answer and the original question): I believe that some of the misunderstandings here and at stackexchange arise from Delgado missed the implicit assumption point that $L/K$ is M=Fix(Aut_M(L))$isn't a finite extension, which formal consequence of course is not a requirement for being Galois. Still, if$L/K$is K=Fix(Aut_K(L))$ for algebraic and $K$ is the fixed field of $Aut_K(L)$, then $L/K$ is Galois in the usual sense. If extensions $K\subseteq M\subseteq L$. I took a closer look into the (master?) thesis. It doesn't claim to contain anything new, then $Gal(L/M)$ This work is a closed subgroup journey through the main ideas and sucessive [sic] generalizations of Galois Theory, towardsthe profinite group $Gal(L/K)$, so by origins of Grothendieck’s theory of Dessins d’Enfants ... as the usual infinite author puts it in his abstract.

The chapter on Galois theory $M$ is just repeats well-known text book material, mostly without proofs. Considering the fixed field verbose character of $Gal(L/M)$, so $L/M$ is Galois in this other sense toochapter, I'm sure the author would have said more than We immediately conclude that .

All .. if there had been a novel aspect. To me it appears that he simply missed an essential aspect of this really requires Galois theory.

At any rate, from $K=Fix(Aut_K(L))$ alone we cannot conclude much, one somehow has to use the fact that $L/K$ is algebraic . For if too as the following example shows: If $L=K(x)$ for a transcendental $x$, and if $K$ is infinite, then again $K$ is the fixed field of $Aut_K(L)$, but for most rational functions $r(x)$ the extension $L/K(r(x))$ isn't Galois in either sense.

Added (in response to Martin's comment): I still believe that Delgado is a little quick here. However, the ingredients

So if we want to get that $L/M$ is Galois are contained in his thesis too: After his remark he proves that $L/K$ is normal and separable, and before his remark he proves show that each $K$-homomorphism of a subfield of $L$ extends to M=Fix(Aut_M(L))$for an automorphism of$L$. That's all we need: Let$a\in L$be fixed under$Gal(L/M)$, and let$f\in M[x]$be the minimal polynomial of algebraic extension$a$over L/K$ with $M$. Note K=Fix(Aut_K(L))$, then I believe that$f$splits into linear factors over$L$by normality of$L/M$. Let$b$be another root of$f$. So there one is a$K$-homomomorphism$K(b)\to K(a)$mapping$a$to$b$which extends automatically lead to an automorphisms of the normal extension$L$. So, as$a$is fixed by$Gal(L/M)$, we get$f(x)=x-a$, so$a\in M$. In summary, it seems to me that there is no new treatment usual kind of Galois theory herearguments, only the order of which by the arguments might be a little differentare also listed in this thesis. 2 added 940 characters in body I believe that some of the misunderstandings here and at stackexchange arise from the implicit assumption that$L/K$is a finite extension, which of course is not a requirement for being Galois. Still, if$L/K$is algebraic and$K$is the fixed field of$Aut_K(L)$, then$L/K$is Galois in the usual sense. If$K\subseteq F\subseteq M\subseteq L$, then$Gal(L/F)$Gal(L/M)$ is a closed subgroup of the profinite group $Gal(L/K)$, so by the usual infinite Galois theory $F$ M$is the fixed field of$Gal(L/F)$, Gal(L/M)$, so $L/F$ L/M$is Galois in this other sense too. All of this really requires that$L/K$is algebraic. For if$L=K(x)$for a transcendental$x$, and if$K$is infinite, then again$K$is the fixed field of$Aut_K(L)$, but for most rational functions$r(x)$the extension$L/K(r(x))$isn't Galois in either sense. Added (in response to Martin's comment): I still believe that Delgado is a little quick here. However, the ingredients to get that$L/M$is Galois are contained in his thesis too: After his remark he proves that$L/K$is normal and separable, and before his remark he proves that each$K$-homomorphism of a subfield of$L$extends to an automorphism of$L$. That's all we need: Let$a\in L$be fixed under$Gal(L/M)$, and let$f\in M[x]$be the minimal polynomial of$a$over$M$. Note that$f$splits into linear factors over$L$by normality of$L/M$. Let$b$be another root of$f$. So there is a$K$-homomomorphism$K(b)\to K(a)$mapping$a$to$b$which extends to an automorphisms of the normal extension$L$. So, as$a$is fixed by$Gal(L/M)$, we get$f(x)=x-a$, so$a\in M$. In summary, it seems to me that there is no new treatment of Galois theory here, only the order of the arguments might be a little different. 1 I believe that some of the misunderstandings here and at stackexchange arise from the implicit assumption that$L/K$is a finite extension, which of course is not a requirement for being Galois. Still, if$L/K$is algebraic and$K$is the fixed field of$Aut_K(L)$, then$L/K$is Galois in the usual sense. If$K\subseteq F\subseteq L$, then$Gal(L/F)$is a closed subgroup of the profinite group$Gal(L/K)$, so by the usual infinite Galois theory$F$is the fixed field of$Gal(L/F)$, so$L/F$is Galois in this other sense too. All of this really requires that$L/K$is algebraic. For if$L=K(x)$for a transcendental$x$, and if$K$is infinite, then again$K$is the fixed field of$Aut_K(L)$, but for most rational functions$r(x)$the extension$L/K(r(x))\$ isn't Galois in either sense.