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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.

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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.
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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.