Galois correspondence counter-example

  • Thread starter PsychonautQQ
  • Start date
In summary: The Attempt at a SolutionIn summary, the Galois correspondence links intermediate fields to subgroups of the Galois group. The map that takes an intermediate field to it's subgroup and " the map that takes a Galois subgroup to it's intermediate field are both Galois maps, but there is only one map that satisfies the relation between the two, which is the map that takes a Galois subgroup to it's Galois group.
  • #1
784
10

Homework Statement


Hey PF! Alright so the Galois correspondence links intermediate fields to subgroups of the Galois group. Let's 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.

Homework Equations

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.
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes PsychonautQQ
  • #3
fresh_42 said:
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\,##?

##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
PsychonautQQ said:
##H \leq G## belonging to ##L##, does not make sense, as L is an intermediate field ...
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.
 
  • Like
Likes PsychonautQQ
  • #5
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
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
PsychonautQQ said:
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.
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.
 
  • Like
Likes PsychonautQQ

1. What is Galois correspondence and how does it work?

Galois correspondence is a mathematical concept that relates algebraic structures, such as fields and groups, to the substructures contained within them. It states that for every field extension, there is a corresponding subgroup of its Galois group, and vice versa. This allows for a deeper understanding of the relationships between these structures and can be used to solve various problems in algebra.

2. What is a counter-example in Galois correspondence?

A counter-example in Galois correspondence is a specific case where the correspondence does not hold. In other words, it is an example where the expected relationship between a field extension and its corresponding Galois group subgroup does not exist. This can be a useful tool in understanding the limitations of Galois correspondence.

3. How can a counter-example be used in Galois correspondence?

Counter-examples can be used to illustrate the nuances and limitations of Galois correspondence. By studying counter-examples, mathematicians can gain a deeper understanding of the theory and potentially find ways to improve it. They can also be used as a way to test the validity and applicability of Galois correspondence in different scenarios.

4. What are some common counter-examples in Galois correspondence?

One common counter-example in Galois correspondence is the case of non-normal extensions, where the correspondence does not hold. Another is the case of infinite Galois groups, where the correspondence becomes more complicated. Additionally, there are many specific examples in different algebraic structures that can serve as counter-examples.

5. Can a counter-example disprove Galois correspondence?

No, a single counter-example cannot disprove the entire concept of Galois correspondence. It is important to remember that counter-examples are specific cases and do not necessarily represent the overall trend or validity of the theory. However, they can provide valuable insights and help refine our understanding of Galois correspondence.

Suggested for: Galois correspondence counter-example

Replies
1
Views
597
Replies
1
Views
862
Replies
6
Views
1K
Replies
15
Views
1K
Replies
9
Views
931
Back
Top