# Galois correspondence counter-example

1. Dec 1, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Hey PF! Alright so the Galois correspondence links intermediate fields to subgroups of the Galois group. Lets call * the map that takes an intermediate field to it's subgroup and " the map that takes a Galois subgroup to it's intermediate field. I need to find an example for a subgroup H such that H"* does not equal H. H will always be included in H"* but will only equal H under some special circumstances that I don't completely understand, I don't think it's as easy as if the field extension is Galois.

2. Relevant equations

3. The attempt at a solution
I've tried quite a few things. I started off with something basic like Q(a,c):Q where a is a 3rd cube root of 2 and c is a primitive fourth root or something like that. I figure my best luck will be if I look for extensions that aren't galois. Then I tried intermediate fields Q(ac), Q(a), Q(c) but couldn't find anything who's subgroup it was mapped to had the property I was looking for, which is again H"* not equalling H. If anyone has any tips for me that'd be awesome.

2. Dec 1, 2016

### Staff: Mentor

I am a bit confused and trying to find out, whether it is mine or yours. Let us define some fields and groups, to find a common language.
First we start with a Galois extension of fields, say $K \subseteq M$. It's Galois group $G=Gal(M:K)$ is the group of all automorphisms of $M$ which leave the elements of $K$ fixed. We further have a subgroup $U \leq G$.

Now (in my book) second theorem of Galois theory tells us, that there is exactly one intermediate field $K \subseteq L \subseteq M$, such that $U = Gal(M:L)$, all automorphisms of $M$ which fix $L$. In your (by the way hard to read choice of mapping names) $U"=L$.
We know further that $[M:L]=|U|$ and $[L:K]=|G/U|$.

Here comes where you lost me:
According to (in my book) first theorem of Galois theory, there is a subgroup $H \leq G$ belonging to $L$, such that $H=Gal(M:L)$, with your mappings $H=L^*$. Why isn't $H=Gal(M:L)=U$, i.e. $H=L^*=(U")^*=U\,$?

So we have the following possibilities, as I see it:
• I might have read something wrong, for it's late here and my book is printed with rather small letters.
• You see, where we differ and can clarify the situation.
• Maybe you could tell, what it was, that you didn't "completely understand".
• We both hope someone smarter than us will help us here.
What I suspect is, that we are talking about normality: the correspondence between normal subgroups and normal field extensions.
I have mentioned above, that $[L:K]=|G/U|$. Now, $G/U$ in general isn't a subgroup of $G$, i.e. the correspondence fails on the lower (left) part. But this is more of a wild guess rather than understanding what you were looking for.

3. Dec 2, 2016

### PsychonautQQ

$H \leq G$ belonging to $L$, does not make sense, as L is an intermediate field and H is a subgroup of $Gal(M:K)$. I will assume that you meant to say $L*$, which is the subgroup of $Gal(M:K)$ such that every automorphism fixes $L$ point wise. But if this is what you meant by $L*$ then $L*=Gal(M:L)=H$.

Here is exactly word for word the question I am trying to answer:

1.) Let L:K be a field extension. Let A be the set of subgroups of the Galois group Gal(L:K) and B be the set of intermediate fields of the extension L:K. Find an example of a subgroup H of a Gal(L:K) such that H is not equal to H"* (H will still be contained in H"* as I proved in an earlier exercise).

So yeah, they really leave a lot of room open for us. We get to define what L and K are and then try to find a subgroup of the Galois group such equality doesn't hold when you apply the map that takes it to it's intermediate field, and then the intermediate field you arrived at back to a subgroup.

4. Dec 2, 2016

### Staff: Mentor

Maybe "belonging" is the wrong word, but I meant $L$ to be the intermediate field, such that $H$ is its Galois group.

I'm still puzzled as I thought it would be a 1:1 correspondence between $A$ and $B$ leading back and forth.
I'll translate what van der Waerden has proven in his book. Perhaps we or someone else can see the discrepancy and work it out. At least it might give us a hint, how to find a counterexample. Notation changed in your quote (cp. 1.)), so we'll have to agree on one. I'll change towards $K \subseteq M \subseteq L$ according to your usage, i.e. $H \leq G=Gal(L:K)$ and $M$ as in interMediate. Unfortunately, van der Waerden's language is a bit old fashioned, which might well be the source of my confusion. So here we go:

1. To each intermediate field $K \subseteq M \subseteq L$ belongs a subgroup $H$ of $G$, namely all automorphisms of $L$, which leave all elements of $M$ fixed.
2. $M$ is uniquely determined by $H$; as $M$ consists of all elements of $L$, that allow the 'substitutions' of $H$, i.e. stay invariant under $H$.
3. To each subgroup $H$ of $G$ there can be found a field $M$, which is related to $H$ as described above.
4. The order of $H$ equals the degree $[L:M]$; the index of $H$ in $G$ is equal to the degree $[M:K]$.

Now here is how I read this: (I used two names $U,H$ for the subgroups, in order to distinguish them w.r.t. their logical quantor.)

1. $\forall M \; \exists U \, : \,M^* = U \leq G$
2. $M = U"$
3. $\forall H \leq G \; \exists M\, : \, H" = M$
4. $|H|=[L:M]\, , \, |G/H|=[M:K]\, , \,|G| = [L:K]$

So if we start with a subgroup $H \leq G$, then 3. guarantees us an intermediate field $K \subseteq M_H \subseteq L$ with $H"=M_H$. 2. simply says that $":A \rightarrow B$ is injective. Now by 1. we find a subgroup $U_M \leq G$ to our field $M_H$ such that $U_M=M_H^*$.

Thus $U_M=M_H^*=(H")^*$ and you said that $H \subseteq (H")^*$. Hence we are looking for an example $H \subseteq U_M$.
But my understanding of 4. says $|U_M|=[L:M_H]=|H|$, hence the inclusion cannot be proper, i.e. $H = U_M = (H")^*$.

Now where is my mistake? Maybe I've overlooked something obvious and got caught in a logical circle, this cannot be ruled out.

5. Dec 2, 2016

### PsychonautQQ

So I asked my professor and it turned out everything you're saying is correct and it was a bad question.... But like usual your post was very enlightening even though it was spurred by a faulty question! Thank you a ton!

6. Dec 2, 2016

### PsychonautQQ

Are your points 1-4 dependent on the extension being Galois? Not that it matters anymore, but I was looking for a counter example in any extension, not necessarily a Galois.

7. Dec 2, 2016

### Staff: Mentor

I've asked myself the same question. The alternative would be, that $L$ is not a splitting field of $K$. But then $Aut(L:K)$ isn't a Galois group and our correspondence isn't proven any longer.

We've had $\mathbb{Q} \subseteq \mathbb{Q}(c)$ with $(c^3-2)$ as an example for such an extension in another thread. Let's say we change this to $x^{6}-2=0$ for a larger set of roots to have the chance of some subgroups. What is $Aut(\mathbb{Q}(c):\mathbb{Q})$ for one $c$ with $c^6=2$ then? I guess we will sooner or later have to consider its splitting field which will find us in the situation of a Galois group / extension again, since $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[6]{2})$ won't lead very far.