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B Why does every subfield of Complex number have a copy of Q?

  1. Jun 11, 2017 #1
    Why does every subfield of Complex number have a copy of rational numbers ?

    Here's my proof,

    Let ##F## is a subield of ##\Bbb C##. I can assume that ##0, 1 \in F##.
    Lets say a number ##p \in F##, then ##1/p \ p \ne 0## and ##-p## must be in ##F##.

    Now since ##F## is subfield of ##\Bbb C## it has the operation of mutiplication and addition.

    So for some ##q \in F## there exist ##q/p## and ##p/q##( ##p,q \ne 0## ) in ##F##. Therefore every number of form ##p/q, q \ne 0## exist in ##F##.

    Is this enough ?
     
  2. jcsd
  3. Jun 11, 2017 #2

    FactChecker

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    If your goal is to learn how to do formal math proofs, here are a couple of comments:
    Is there something in the problem statement that rules out the trivial subfield {0}?
    "assume" should raise a red flag. You should prove this.
    You should start with an arbitrary element of Q and show that it is in F.
     
    Last edited: Jun 11, 2017
  4. Jun 11, 2017 #3

    fresh_42

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    As already mentioned, I would replace "assume" by an argument why this must be true, too. Besides this, the only question open is, why can't it happen that you get cycles in ##F##, i.e. do you obtain all quotients or might it happen that for ##p \in F## you get ##p \cdot n \in \{0,1\}## which would prevent you from getting all numbers you need. Why is this impossible? And why is there a ##p \in F## at all? ##0## and ##1## are basically clear, but why are there other elements? ##\mathbb{Z}_2## is also a field. Can it be a subfield of ##\mathbb{C}## ?
     
  5. Jun 11, 2017 #4
    Original question : upload_2017-6-11_21-3-43.png

    I think the author assumes that we are talking about non trivial fields.


    I assumed ##0,1## because the definition of field given in book says that a field must contain additive and multiplicative inverse.

    What does ##p \cdot n ## mean ?
    Is ##\Bbb Z_2## same as ##\Bbb Z^2## ?
     
  6. Jun 11, 2017 #5

    fresh_42

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    There is no such thing as a trivial field. Every field of characteristic zero contains the rational numbers. It is called the prime field of characteristic zero.
    Yes. Only the wording has been strange. You do not assume, you know that ##0,1## are in any field.
    It means ##p+p+ \ldots + p## , ##n-## times.
    No. ##\mathbb{Z}_2 = \{0,1\}##, the smallest possible field. ##\mathbb{Z}^2 = \mathbb{Z} \times \mathbb{Z}## is a ring and not a field.

    You have to find an argument to get all rational numbers. But addition only gets you all, if there are no cycles, no numbers that turn to zero by addition like in ##\mathbb{Z}_2## in which ##1+1=0##. So why are all elements ##\frac{1}{p},\frac{2}{p}, \frac{3}{p}, \ldots ## different? It's not really hard to say why, it simply has to be mentioned. ##\mathbb{Z}_2 = \{0,1\} \subseteq \mathbb{C}## as a set, but not as a field. Where's the difference? And there is still an argument missing, why e.g. ##p \in F - \{0,1\}##. Again nothing complicated, just a short reason for it.

    You can start with ##0## and ##1## which you have and then construct first all naturals (half group), then all integers (group, integral domain) and at last all quotients (field). There must be a reason why ##\mathbb{Z}_2## isn't the smallest subfield, which?
     
  7. Jun 11, 2017 #6

    WWGD

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    Start with ## 1##. Then ##1+1 ## is in ##F##, and so is ## \frac {1} {1+1} ## .....
     
  8. Jun 11, 2017 #7

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    Sorry. I stand corrected. The inclusion of different additive and multiplicative identity elements (0 and 1) is one of the field axioms. So @fresh_42 is correct -- there is no "trivial" field {0}.
     
    Last edited: Jun 11, 2017
  9. Jun 11, 2017 #8
    Ok, then ##\Bbb Z_2## cannot be a subfield because a subfield of complex number has characteristics zero and ##\Bbb Z_2## has a characteristics of 2.



    Ok here is my proof again.

    Lets start with ##1##. Since ##1 + 1## cannot be zero, so we introduced a new element ##2##.
    Next we to ##2 + 1## this also should not be zero because so we introduced element ##3##, and so on till be get all natural number.

    There also exist multiplicative and additive inverse of each element in ##F##. Since we already know that any natural number added to any other won't give ##0##, so we introduced new elements ##-1, -2, -3, \cdots ## corresponding to ##1, 2, 3\cdots ##.
    Similarly we get multiplicative inverses ##... -1/2, -1/1 , 1/1, 1/2 ...## for each element in ##F##.

    Therefore we have got ##1/q## ( ##q \ne 0## ) in ##F## where ##q \in \Bbb Z##.
    To get ##p/q## where ##p \in \Bbb Z## we just multiply ##p## and ##1/q##.

    Thus we got every element of form ##p/q## where ##p,q \in \Bbb Z## and ##q \ne 0##.

    Puff, is this correct now ?
     
  10. Jun 11, 2017 #9

    fresh_42

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    Yes. Only a minor remark: "because ##\operatorname{char} \mathbb{C} = 0## and a subfield cannot have another" provokes the question why? The reason is, that an equation ##1+ \ldots +1 = 0## in the subfield ##F## is also an equation in ##\mathbb{C}## and therefore it is impossible. I know it is kind of hair splitting, but it is not trivial, because it uses the fact, that the embedding ##F \subseteq \mathbb{C}## is injective. If you consider it trivial, then the entire exercise is.
     
  11. Jun 11, 2017 #10
    Oh right I did not knew the proof of the that fact. I just took as stated thinking it would be very difficult to prove.

    Is the converse also true ? that each field with ##\operatorname {char} 0## is a subfield of complex numbers ?
     
  12. Jun 11, 2017 #11

    fresh_42

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    No, this is wrong. The simplest example would be the field of rational functions with rational (or real) coefficients: ##\mathbb{Q}(x)##. It is the quotient field of the polynomial ring ##\mathbb{Q}[x]##, that is it contains all quotients of polynomials in ##x##. This field is obviously not contained in ##\mathbb{C}##. But all fields of characteristic ##0## contain the rational numbers, as you have just proven, because you haven't used any property of complex numbers beside the fact, that ##1+ \ldots +1 \neq 0## no matter how many ##1##'s we add.

    Another not as simple example are the p-adic numbers with very unusual properties.
     
  13. Jun 11, 2017 #12

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    I think that worrying about cycles shows that you are taking a confusing approach to the proof. It may be just the wording that seems strange to me. You only need to show that F contains Q. So pick an arbitrary p/q number in Q and show that it is in F.
     
  14. Jun 12, 2017 #13
    Let ##p/q## be in ##\Bbb Q##. Since ##p/q = p\times 1/q = (1 + 1 + \stackrel{\text{p times}}{\cdots} + 1) \times 1/(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)##.

    Since ##1,0## is there in ##F## and ##F## has a characteristic of ##0## therefore both ##(1 + 1 + \stackrel{\text{p times}}{\cdots} + 1)## and ##(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)## is in ##F##.

    Done.
     
  15. Jun 12, 2017 #14

    WWGD

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    For being finicky's sake:

    Given ## p/q; q \neq 0 ##:
    a)##p,q := 1+1+1+.....## ( ##p, q## times respectively ; it is not 0 , since the Complexes have characteristic 0. )We have p,q
    b) Every element in a field is invertible; in particular, q is invertible, so ## 1/q## is in ## F##
    c) By closure of operations, ## p (1/q):=p/q ## is in ##F##.
     
  16. Jun 12, 2017 #15

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    @WWGD is not just being "finicky". If you are learning how to do math proofs, remember that proofs should be done in methodical "baby steps". Only the most trivial proofs can be done otherwise. Try to make every statement depend on one fundamental fact, that you state. You can start to skip and combine steps after you have a lot of experience.
     
  17. Jun 12, 2017 #16

    WWGD

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    Thanks, it is just that one gets often "accused" of this when being (necessarily , I agree) careful.
     
  18. Jun 12, 2017 #17

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    The fact that 1+1+...+1 (p times) is in F is due to closure under addition, not due to the characteristic. The sum is in F regardless of whether it cycled back to 0 or not. Similariy for q. You only need the characteristic to say that q≠0 and therefore 1/q is in F.
     
  19. Jun 12, 2017 #18

    fresh_42

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    Strictly speaking you also need the injectivity of ##F## in ##\mathbb{C}## to preserve the characteristic which is needed to get all rationals. So if it cycles back it is still in ##F## but that would not be enough. But you (both) are right: this is a good example to see that many properties of ##F## are actually used.
     
  20. Jun 12, 2017 #19

    WWGD

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    When would you not have an injectivity, other than for a finite field?
     
  21. Jun 12, 2017 #20
    No I said about the characteristic of ##F## because if it cycles back to 0, then there is at least one rational number which is not in field.
    Closure of addition is an axiom so I took it for granted.
     
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