Generally a Galois extension is defined to be an algebraic extension that is also normal & separable. It is then shown that in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This follows since the same property is valid for separable & normal extensions individually. It also follows that $L|K$ is a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$

In Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory he takes the opposite tack, and in definition defines an algebraic extension of fields $L|K$ to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$.

It is then shown that in the sequence of algebraic field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction (and so has no details), I don't see the obviousness...can someone clarify.

In proposition he shows that Galois extension is normal and separable.

All this appears to be in the opposite order of the standard treatments. One reason I'm interested in his formulation, if it is correct, is that one side of the Galois correspondence follows easily from this.

disclaimer: I've already asked this question on math.stackexchange but the answers there revolved around characterising Galois extensions as being normal & separable, and then showing this property follows.

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    $\begingroup$ I can't see how the question wasn't answered in the Stackexchange thread. Try reading it again, and consulting your Galois theory textbook? $\endgroup$ Jan 31, 2013 at 14:44
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    $\begingroup$ The question as I read it seems really not answered on math.SE, possiblty because it was misunderstood. OP does not seem to ask how one can proof this at all but rather: Suppose we define an extension to be Galois if the field fixed under Aut_K(L) is K. Is there then an 'obvious' reason that for an intermideate field M also the extension L over M is Galois. [I am not sure this is an appriate question ATM; an have no time to decide, but in any case I feel the question is partly misunderstood.] $\endgroup$
    – user9072
    Jan 31, 2013 at 17:45
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    $\begingroup$ This is an interesting question, which should be read more carefully (especially by the ones who down/close voted it because it seems to be elementary). Here a reformulation: If $L/M/K$ are algebraic field extensions, and $K = L^{\mathrm{Aut}_K(L)}$, how can we prove directly that $M = L^{\mathrm{Aut}_M(L)}$? $\endgroup$ Jan 31, 2013 at 18:22
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    $\begingroup$ @Martin & Mozibur. (Assuming $G=\mathop{\rm Aut}_K(L)$ finite). It is obvious that $M$ is contained in $L^{\mathop{\rm Aut}_M(L)}$, what needs to be shown is the other inclusion. Or: if $x\not\in M$, there exists $g\in\mathop{\rm Aut}_M(L)$ such that $g(x)\neq x$. Now, all conjugates of $x$ over $M$ belong to $L$, and one of them, say $y$, is distinct from $x$. This gives an $M$-linear morphism $M[x]\to L$ such that $x\mapsto y$. Going on, this morphism can be extended to a $M$-linear morphism from $L$ to $L$, which is then an element $g$ of $G$ such that $g(x)=y$. $\endgroup$
    – ACL
    Feb 1, 2013 at 12:02
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    $\begingroup$ @ACL: You use that $L/M$ is normal and separable. $\endgroup$ Feb 1, 2013 at 15:52

2 Answers 2


Edited (in view of new comments to this answer and the original question): I believe that Delgado missed the point that $M=Fix(Aut_M(L))$ isn't a formal consequence of $K=Fix(Aut_K(L))$ for algebraic extensions $K\subseteq M\subseteq L$. I took a closer look into the (master?) thesis. It doesn't claim to contain anything new, This work is a journey through the main ideas and sucessive [sic] generalizations of Galois Theory, towards the origins of Grothendieck’s theory of Dessins d’Enfants ... as the author puts it in his abstract.

The chapter on Galois theory just repeats well-known text book material, mostly without proofs. Considering the verbose character of this chapter, I'm sure the author would have said more than We immediately conclude that ... if there had been a novel aspect. To me it appears that he simply missed an essential aspect of 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 too as the following example shows: If $L=K(x)$ for a transcendental $x$, and if $K$ is infinite, then $K$ 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.

So if we want to show that $M=Fix(Aut_M(L))$ for an algebraic extension $L/K$ with $K=Fix(Aut_K(L))$, then I believe that one is automatically lead to the usual kind of arguments, which by the are also listed in this thesis.

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    $\begingroup$ Once again, this doesn't answer the question, and probably Mozibur knows all that (see also the math.SE discussion). Mozibur has asked if there is a proof which avoids the usual characterization of Galois extensions as well as the Main Thm on Galois theory, because Robalo Delgados indicates that this is possible. And even for finite extensions this is an interesting question. $\endgroup$ Jan 31, 2013 at 19:35
  • $\begingroup$ You show that $a$ is the unique root of $f$, but this only implies that $f(x)=x-a$ if we already knew that $L/M$ is separable. You also use in the proof that $L/M$ is normal. Therefore, again this is just the proof (reducing to the statements for normal and separable) which Mozibur wants to avoid. Of course, I don't claim that this is possible at all, but it would be interesting. $\endgroup$ Feb 1, 2013 at 0:14
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    $\begingroup$ My comments refer to older versions of the answer. $\endgroup$ Feb 1, 2013 at 10:01
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    $\begingroup$ @Mueller: I'm beginning to suspect that Delgado is wrong in his claim, particularly the 'immediacy' of the deduction... $\endgroup$ Feb 1, 2013 at 13:41

This proof might not count as direct. It is for finite extensions.

Lemma: $L/K$, a finite extension, is Galois if and only if $L \otimes_K L$ is a product of copies of $L$.

Proof: If $L\otimes_K L$ is a product of copies of $L$, and $x$ is fixed by every automorphism, then $s \otimes 1- 1 \otimes s$ is zero in $L \otimes_K L$, so $s \in K$ (via explicit description of tensor products of vector spaces.)

If there are a lot of automorphisms, then each automorphism gives a different surjective map $L \otimes_K L \to L$, so we get a surjective map from $L \otimes_K L $ to the product of $[L:K]$ copies of $L$, which must be an isomorphism by dimension-counting.

Then $L \otimes_M L$ is a quotient of $L \otimes_K L$, so is a product of finitely many copies of $L$.

I believe one can extend this to all algebraic extensions via a slightly more complicated argument. But that might just be pointless, as one could argue that my condition is just a clever way of saying "normal and separable" in different language.

  • $\begingroup$ Dear Will, when you say <<$L \otimes_K L$ is a product of copies of L>>, by product you mean for example $L \times L$? Thanks. $\endgroup$ Feb 1, 2013 at 3:23
  • $\begingroup$ Yes. Obviously it being a tensor product does not say much. $\endgroup$
    – Will Sawin
    Feb 1, 2013 at 3:31
  • $\begingroup$ Is it obvious that if there are "enough" automorphisms then there must be $[L:K]$ of them? $\endgroup$ Feb 1, 2013 at 14:23
  • $\begingroup$ I can't think of an easy proof. There is a proof that is standard in Galois theory, but I think that just makes this the regular proof in new clothing. $\endgroup$
    – Will Sawin
    Feb 1, 2013 at 16:41

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