Uniqueness of ordering of a ordered field.

In summary, the conversation discusses the concept of uniqueness in ordered fields and whether it is possible for two different orderings to make a field into an ordered field. The conversation explores various attempts at finding a contradiction to prove uniqueness, including using rational numbers and constructing counterexamples. Ultimately, the conversation concludes that the uniqueness of ordering in a field depends on the field operations and cannot be determined solely based on the elements within the field.
  • #1
FallMonkey
11
0

Homework Statement



Is the ordering of a ordered field unique? That is, is it possible to have different ordering set(the order), we call P1 and P2, both able to make a field F into a ordered field?

Homework Equations



no.

The Attempt at a Solution



First I tried to assume now there's a X in P1, but not in P2(because P1 != P2), and therefore by definition -X must be in P2. After that, I tried to get to a point where I can make X also in P2 to have contradiction. But I can't.

Then I tried to go from ration numbers. With some existing theorem, Q(rational number set) is the smallest ordered field, and a subfield of any ordered field. Therefore if there're such P1 and P2 in some ordered field, they will both have Q, thus the element that is not shared can not be a rational number. After that, I stopped without any idea how i could progress.

Thank you very much for viewing, and let me know anything I unclearly stated.
 
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  • #2
Think about extension fields of Q. Like Q[sqrt(2)]. Is there anything in the ordered field properties that forces sqrt(2) to be in P?
 
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  • #3
Dick said:
Think about extension fields of Q. Like [Q,sqrt(2)]. Is there anything in the ordered field properties that forces sqrt(2) to be in P?

By applying taylor series to express sqrt(2) as result of a series of rational numbers I guess?

Therefore any irrational number must be in P...and then we could get to real number field. But it's still not a general field, and I can't prove the uniqueness of its order.

It's a real analysis class and I'm so not sure of what I can use and what I can't, because everything I know for sure is getting proved.
 
  • #4
FallMonkey said:
By applying taylor series to express sqrt(2) as result of a series of rational numbers I guess?

Therefore any irrational number must be in P...and then we could get to real number field. But it's still not a general field, and I can't prove the uniqueness of its order.

It's a real analysis class and I'm so not sure of what I can use and what I can't, because everything I know for sure is getting proved.

Any irrational number must be in P? Now that's gibberish. I'm suggesting you use Q[sqrt(2)] to construct a counterexample. There are two field orderings of Q[sqrt(2)].
 
  • #5
Dick said:
Any irrational number must be in P? Now that's gibberish. I'm suggesting you use Q[sqrt(2)] to construct a counterexample. There are two field orderings of Q[sqrt(2)].

Ehh sorry my bad understanding and stupidity.

So for Q[sqrt(2)], either sqrt(2) is in P or -sqrt(2) is in P. And we know all a+bsqrt(2) will be in Q.

by constructing two field ordering of Q[sqrt(2)], and by proving that sqrt(2) is always in P, we have the contradiction, therefore the uniqueness, right?

I come up with something like a+bsqrt(2) where b is postive is in P, and a+bsqrt(2) where b is negative is in P, thus two different ordering but both satisfying the requirement of ordering. And by fact that sqrt(2) is always there, we have the contradiction?

My approach still sounds not flawless. Am I wrong at some steps this time?
 
  • #6
FallMonkey said:
Ehh sorry my bad understanding and stupidity.

So for Q[sqrt(2)], either sqrt(2) is in P or -sqrt(2) is in P. And we know all a+bsqrt(2) will be in Q.

by constructing two field ordering of Q[sqrt(2)], and by proving that sqrt(2) is always in P, we have the contradiction, therefore the uniqueness, right?

I come up with something like a+bsqrt(2) where b is postive is in P, and a+bsqrt(2) where b is negative is in P, thus two different ordering but both satisfying the requirement of ordering. And by fact that sqrt(2) is always there, we have the contradiction?

My approach still sounds not flawless. Am I wrong at some steps this time?

That's not really it. Think about this. Q[sqrt(2)] is the set of all numbers a+sqrt(2)*b with a and b rational. There is a field automorphism that exchanges sqrt(2) and -sqrt(2) sort of like complex conjugation, right? If you decide to put sqrt(2) in P, then you don't have much choice where to put it in the ordering of Q. It has to go between 1 and 2 in the usual 1.414... spot. But you could also have put -sqrt(2) there. You can't tell the difference between the two using field operations.
 
  • #7
Dick said:
That's not really it. Think about this. Q[sqrt(2)] is the set of all numbers a+sqrt(2)*b with a and b rational. There is a field automorphism that exchanges sqrt(2) and -sqrt(2) sort of like complex conjugation, right? If you decide to put sqrt(2) in P, then you don't have much choice where to put it in the ordering of Q. It has to go between 1 and 2 in the usual 1.414... spot. But you could also have put -sqrt(2) there. You can't tell the difference between the two using field operations.

Oh good morning, wish you a nice dream.

yes I do have no way of telling the difference between the two using field operations, just like that I can't tell the difference between P1 with Z in it and P2 with -Z in it. Is that breaking the uniqueness? I thought I could find a contradiction and prove the uniqueness. Ahh, I can't even understand how you go from there to the proof.

Sorry I'm just starting set theory and abstract math this month. There's too much I don't know and learning day by day. That must be causing toughness for you to guide me through it, though I see you're trying to teach me the way not the answer. PF is full of good guys woot.
 
  • #8
Are you still there, Dick...m(-_-)m
 
  • #9
FallMonkey said:
Are you still there, Dick...m(-_-)m

Sure. I wasn't sure if you still had a question. All the field Q knows about sqrt(2) or -sqrt(2) is that they satisfy x^2=2. That's all. You can either make the standard order where you put sqrt(2) in between 1 and 2 where it belongs, or you can make a nonstandard order where you put -sqrt(2) there instead. The rationals don't care.
 
  • #10
Dick said:
Sure. I wasn't sure if you still had a question. All the field Q knows about sqrt(2) or -sqrt(2) is that they satisfy x^2=2. That's all. You can either make the standard order where you put sqrt(2) in between 1 and 2 where it belongs, or you can make a nonstandard order where you put -sqrt(2) there instead. The rationals don't care.

Ouch, so the uniqueness I'm trying to find is actually not there? Ahh the theorem of real closed field with unique ordering must be understood terribly be me.

I think for one day I totally complicated the problem too far, while at first guess that P1,P2,+Z,-Z(which I can't prove to be wrong, and actually correct if I make Z a unrelated thing to all members left). Thx bro, you helped a lot.

Time for consuming your replies and get every new words there wikified!

Good night.
 
  • #11
FallMonkey said:
Ouch, so the uniqueness I'm trying to find is actually not there? Ahh the theorem of real closed field with unique ordering must be understood terribly be me.

I think for one day I totally complicated the problem too far, while at first guess that P1,P2,+Z,-Z(which I can't prove to be wrong, and actually correct if I make Z a unrelated thing to all members left). Thx bro, you helped a lot.

Time for consuming your replies and get every new words there wikified!

Good night.

Show this, ok? If F is a field, f:F->F is a field isomorphism and P is a field ordering then f(P) is ALSO a field ordering. It's pretty easy. That might make what's going on clearer.
 
  • #12
Dick said:
Show this, ok? If F is a field, f:F->F is a field isomorphism and P is a field ordering then f(P) is ALSO a field ordering. It's pretty easy. That might make what's going on clearer.

The problem is I don't know what fied isomorphism is (maybe for the whole semester) and when I asked professor if I needed to know about field isomorphism to solve this question he anwsered no. Only thing I know and related to isomorphism is Peano isomorphism I learned in number system.

I totally get what you're saying, and keep reading stuff about isomorphism and automorphism, trying to get a automorphism from Q[sqrt(2)] to Q[sqrt(2)]. But right now I'm still not capable of doing that, though I know it could be simple as mapping but on field level.
 
  • #13
FallMonkey said:
The problem is I don't know what fied isomorphism is (maybe for the whole semester) and when I asked professor if I needed to know about field isomorphism to solve this question he anwsered no. Only thing I know and related to isomorphism is Peano isomorphism I learned in number system.

I totally get what you're saying, and keep reading stuff about isomorphism and automorphism, trying to get a automorphism from Q[sqrt(2)] to Q[sqrt(2)]. But right now I'm still not capable of doing that, though I know it could be simple as mapping but on field level.

Take f(a+b*sqrt(2))=a-b*sqrt(2). It just exchanges sqrt(2) and -sqrt(2).
 

Related to Uniqueness of ordering of a ordered field.

What is an ordered field?

An ordered field is a mathematical structure that combines the properties of a field (a set of numbers with operations such as addition and multiplication) and an order relation (a way to compare elements in the set). In an ordered field, the order relation is compatible with the field operations, meaning that if a and b are elements in the field and a < b, then a + c < b + c and a * c < b * c for any other element c in the field.

What does it mean for an ordered field to be unique?

The uniqueness of an ordered field refers to the fact that there is only one possible way to define an order relation on the elements of a field that is compatible with the field operations. This means that any two ordered fields with the same elements and operations will have the same ordering of the elements.

What is the role of the order relation in an ordered field?

The order relation in an ordered field is important because it allows us to compare elements and determine which ones are greater than or less than others. This is useful for many mathematical concepts, such as finding the maximum or minimum value of a set of numbers, or determining if a function is increasing or decreasing.

What are some examples of ordered fields?

The most commonly known ordered field is the set of real numbers, where the order relation is the familiar "less than" (<) relation. Other examples of ordered fields include the set of rational numbers, the set of complex numbers, and the set of polynomials with real coefficients.

Are there any ordered fields that are not unique?

No, there are no ordered fields that are not unique. The uniqueness of an ordered field is a fundamental property that is necessary for the field to be well-defined and for mathematical concepts to be consistent. If an ordered field was not unique, it would lead to contradictions and inconsistencies in our mathematical understanding of the field.

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