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An algebraic extension of fields $L|K$ is defined to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$.

Apparently in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction in section 3.2.1 of Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory, I don't see the obviousness...can someone clarify.

What interests me in this characterisation is that it shows one direction of the Galois correspondance, (the other being shown by Artins Lemma) without going through characterising Galois extensions as algebraic, seperable & normal extensions. In fact Delgado shows that this characterisation follows from his definition.

Edit: I've also asked this question on math.overflow as I thought most people misunderstood the question. I'd welcome attempts to formulate it in clearer terms.

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Are you familiar at all with the characterization of Galois extensions as algebraic, normal and separable extensions? –  JSchlather Jan 30 '13 at 22:05
@Schlather: Yes, I have come across that characterisation. Using the definition as above these two characterisations are shown to be equivalent. –  Mozibur Ullah Jan 30 '13 at 22:45
@MoziburUllah: Don't look at the statement of the Fundamental Theorem of Galois Theory, look at its proof. For the relevant part of the standard proof of the FTGT see my second answer down below. –  Barbara Osofsky Feb 6 '13 at 23:13
@Osofsky: I have been looking at its proof. I'm reasonably familiar with it now. I was surprised that Delgado made such a claim as he did - as its already half of the statement FTGT - and I couldn't see its obviousness. I don't believe now he was right in his claim given the number of people who have looked at it in math.overflow. –  Mozibur Ullah Feb 7 '13 at 0:48
@MoziburUllah: It not only looks like the fundamental theorem, it is the fundamental theorem. Normality and separability are simply a restatement of what is used in the proof of the Galois correspondence, namely an irreducible polynomial of degree $d$ in $K[X]$ has $d$ distinct roots in $L$. If you use his definition it is 'obvious' that if you do not have enough roots of some irreducible polynomial you cannot have the Galois correspondence since the index of the fixing group is determined by the number of homomorphisms into. $L$ –  Barbara Osofsky Feb 7 '13 at 14:40
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5 Answers

Go to a text on Galois Theory and look in the index for Fundamental Theorem of Galois Theory. It says that there is a one-to-one correspondence between {subfields between L and K} and {subgroups of $Aut_K L$} given by intermediate field $M$ corresponds to subgroup $G$ of $Aut_K L$ fixing $M$ with inverse $G$ corresponds to fixed field of $G$. Since Galois extension is defined by $K$ is the fixed field of $Aut_K L$, I can understand why someone would call an obvious corollary of a Fundamental Theorem 'obvious', although I try to avoid using that word and would prefer 'by the FTGT'.

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@MoziburUllah I am not going to answer the OP's question directly as it is a psychological rather than a mathematical question. I will answer a very closely related question which I believe is what the OP wishes to know, and let others answer the original question as they see fit.

The Fundamental Theorem of Galois Theory says that if $L\supseteq K$ is a normal and separable extension of fields, and $G$ the group of all $K$-automorphisms of of $L$, then there is a $1-1$ Galois correspondence taking subgroups $H$ of $G$ to their fixed fields and fields $M$ with $L\supseteq M\supseteq K$ to the group of $M$-automorphisms of $L$. Our proposition is the converse of this Fundamental Theorem.

Proposition. Let $L\supseteq K$ be an algebraic extension of $K$, and assume there is a $1-1$ Galois correspondence taking subgroups $H$ of $G$ to their fixed fields and fields $M$ with $L\supseteq M\supseteq K$ to the group of $M$-automorphisms of $L$. Then $L\supset K$ must be separable and normal.

Proof. If $L\supseteq K$ is not separable then some irreducible polynomial must have a multiple root, and this can only happen if ${\mathrm{char}}(K)=p<\infty$ and some nonzero $\alpha\in L\setminus K$ has $\alpha^{p^k}$ separable over $K$ but any automorphism of $L$ that fixes $K[\alpha]$ must fix $\left(\bigcup_n\in p^{-n}K[\alpha]\right)\cap L$. By hypothesis $\left(\bigcup_n\in p^{-n}K[\alpha]\right)\cap L=K[\alpha]$ so $L\supseteq K$ must be separable.

Now let $0\ne\alpha \in L$, and let the minimal polynomial of $\alpha$ over $K$ be $p(X)\in K[X]$. Let the roots of $p$ in some splitting field of $p$ be $\{\alpha\ , \alpha_1\ ,\ \cdots \ ,\ \alpha_{deg(p)}\ \}$ and by Lagrange 150 years before Galois, $p$ is irreducible and the coefficients of monic irreducible $p$ are the elementary symmetric functions of all the (distinct) roots of $p$ so the product $\prod_{i=1}^k \ (X-\alpha_i)$ which has a root in $L$ cannot have coefficients in $K$ unless $k=deg(p)$ so $L\supseteq K$ is normal.


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Another answer which has misunderstood the question. –  Martin Brandenburg Feb 9 '13 at 14:30
I cannot conceive of any other reason for Delgados to use the word 'obvious' other than he knew the simple proof of the converse of the FTGT which concerns concepts covered in almost any study of field theory leading up to the FTGT. So please tell me what I do not understand aboout the question. If one knows that Galois correspondence $\iff$ you have a normal and separabls extension, why does it matter that one side of this equivalence does not mention the other. The only difference that I see is in the minds of the mathematicians approaching the problem. –  Barbara Osofsky Feb 9 '13 at 22:24
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You do not have to rephrase the question. You only have to look at the question the right way and think about the proof of the FTGT. Let $L\supseteq K$ be fields. Then $L$ is a normal separable extension of $K$ iff every irreducible polynomial of degree $d$ in $K[X]$ which has a root in $L$ has $d$ distinct roots in $L$ iff the Galois correspondence holds.

EDIT starts here.

The original answer here gives the missing equivalence. The first iff $\implies$ follows since the polynomial factors as a product of $d$ linear factors in the algebraic closure of $L$, and none of them are multiple by separability and all are in $L$ if one of them is by normality. The reverse implication $\Longleftarrow$ follows from the Lagrange Theorem (roughly 50 years before Galois) that the coefficients of an irreducible polynomial $p$ over any field $K$ are symmetric functions of the roots, and those symmetric functions will lie in the field $K\subseteq L$ iff the product of all of the linear factors of $p$ in some algebraic closure lie in $K$ are used to multiply out to $p$ so if one root of $p$ lies in $L$, and there are $d$ distinct roots, all of the roots must lie in $L$ (normal) and there are no multiple roots (separable). Now if $K\subseteq M\subseteq L$ and $q$ is the minimal polynomial in $M[X]$ of an element $\alpha\in L$, $q$ divides the minimal polynomial of $\alpha$ in $K[X]$ so if the polynomial over $K$ has distinct roots so does $q$ over $M$, and the reverse implication of this last iff is just a restriction to $K$.

My recollection from many years ago is that when I learned the proof of the FTGT as stated in terms of normal + separable, the proof immediately went to the equivalent distinct roots property. Is there any other part of the original question I am missing?

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For $K\subseteq L$ and polynomials in $K[X]$ we have: Not separable $\iff$ some irreducible polynomial has a multiple root in $L$. Normal $\iff$ any irreducible polynomial with one root in $L$ has all roots in $L$. –  Barbara Osofsky Feb 7 '13 at 18:43
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If I may refer you to a very readable, free download of what I think is the best entry point to Galois Theory by Andrew Baker. It will not only answer your question, as Barbara Osofsky did above, but most likely be useful for other such questions. You can also get solutions to its nice problem sets.

You can access it here:


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Let $L \subset K$ be a Galois extension of fields. Let $ L \subset M \subset K$ be an intermediary field. Then we know a priori that $L \subset M$ is separable and $M \subset K$ is Galois. The fact both extensions is separable is straightforward. To see that $M \subset K$ is normal let $p(t) \in M[t]$ be a polynomial taking a root $\alpha \in K$. Let $q(t)$ be the minimum polynomial of $\alpha$ over $L$ since $L \subset K$ is a normal extension we know that $q(t)$ splits over $K[t]$. We also know that $p(t)$ divides $q(t)$ in $M[t]$. In particular this gives that $p(t)$ splits in $K[t]$ since it still must divide $q(t)$ in $K[t]$. Now when $L \subset M$ is a Galois extension is a more subtle question. The fundamental theorem of Galois theory is required to answer this question. It tells that $M$ is normal precisely when the subgroup of $Aut_L(K)$ fixing $M$ is normal in $Aut_L(K)$.

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