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Homework Help: When the Product of Non-Squares is a Square

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    For some field F, if a and b are two elements which are not squares but whose product is a square, show that there is an element k such that [itex] a = k^2 b [/itex].

    3. The attempt at a solution

    This is a boiled down version of what I'm actually trying to show, so the statement above may not actually be correct. If it's not, please let me know!

    I have a few ideas of how to hit this, but the following is the one I like the most:

    If [itex] ab = c^2 [/itex] for some element [itex] c \in F [/itex] then it is sufficient to show that either [itex] a | c [/itex] or [itex] b | c[/itex], since then (without loss of generality) if [itex] a | c [/itex] then there is some constant k such that ak = c in which case [itex] ab = a^2 k^2 [/itex] which implies that [itex] b = k^2 a [/itex]. Now the hard part (for me at least!) is showing that [itex] ab = c^2 [/itex] implies that either a|c or b|c. It seems like this result should be true, but I've been having trouble showing it.

    I start like this. If both a|c and b|c then we are done, hence assume without loss of generality that [itex] b \not| c[/itex]. Here is where I get stuck. Any ideas?
     
  2. jcsd
  3. Jan 15, 2012 #2

    Bacle2

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    Sorry to change your strategy, but , in the factoring of a square, every factor should
    appear an even number of times.

    Then ab=∏xi2ki
     
  4. Jan 15, 2012 #3

    Dick

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    Are you sure you are working in a FIELD? If that's the case, then you can always solve ak=c for k if a is nonzero. k=c*a^(-1).
     
  5. Jan 15, 2012 #4
    Bacle2, I agree entirely. In particular, if we apply this to the case when a and b are not squares it implies that a and b must share a prime factor, since otherwise they would be squares themselves. However, I don't quite see why that helps. Care to elaborate?

    Yes, you are of course correct. I wonder if this would actually work for what I was originally trying to do. I'll have to think about this.

    In any case, I believe that the statement

    "If [itex] ab = c^2 [/itex] then either a|c or b|c"

    is still correct. I would be interested in seeing a proof of this if anyone can suggest one.
     
  6. Jan 15, 2012 #5

    Dick

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    The proof is trivial in a field. Everything divides everything. If you are working over the integers then 4*9=36=6^2. Neither 4 nor 9 divides 6.
     
  7. Jan 15, 2012 #6
    Ah excellent. Thank you.
     
  8. Jan 15, 2012 #7

    Bacle2

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    While Dick gave you a nice answer, my idea was to use the fact that fields are UFD's, so that the product ab has a unique expression in which c must appear twice on the LH side.
     
  9. Jan 15, 2012 #8

    Dick

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    How many fields have you seen that are UFD's??
     
  10. Jan 15, 2012 #9

    Bacle2

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    Every field is a UFD, I'm almost 100%. Do you have a counter?
     
  11. Jan 15, 2012 #10

    Dick

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    The rationals, the reals, the complexes, integers modulo a prime p where p isn't 2, fields of order p^n where n>1, Galois extension fields. Do you want me to go on? Basically all of them. Think about it. What are the primes in the rationals, Q?
     
    Last edited: Jan 15, 2012
  12. Jan 15, 2012 #11

    Bacle2

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    I'm talking about irreducibles.

    If F is a field, then F[X] is a UFD, e.g: http://planetmath.org/encyclopedia/PrimeFactor2.html [Broken]

    This means that every poly. in F[X] has a unique factorization. Now,F embeds trivially

    in F[X], right? Or isn't every f in F an element in F[X]?

    Just use the embedding f--> f+ 0.x+0.x2+....+0xn

    It then follows every f in F has a unique factorization.
     
    Last edited by a moderator: May 5, 2017
  13. Jan 16, 2012 #12

    Dick

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    You seem to be googling instead of thinking. Unique factorization of polynomials is modulo the underlying field F. x^2-1=(x-1)(x+1)=(2x-2)(x/2+1/2). 2=3*(2/3)=5*(2/5). Does that look like unique factorization?
     
    Last edited by a moderator: May 5, 2017
  14. Jan 16, 2012 #13

    Bacle2

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    I was referring to your statement:



    So every field is a UFD, by the definition of UFD, since, trivially, every element of a field is a unit. I just used the link to avoid doing a proof of this fact. Maybe you were
    using the term in relation to this problem/situation which I guess I missed, but, unless otherwise stated, every field is a UFD.

    I just want to add that sometimes I go thru old posts, and I prefer to be maybe overly-finicky in using terminology to avoid picking up things that , however answer
    a question, are not fully correct in the strict sense.
     
    Last edited: Jan 16, 2012
  15. Jan 16, 2012 #14

    Dick

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    Yes, everything is a unit. So it's a trivial UFD. This post is what I've actually been going on about. What is the 'unique expression'? 1?
     
  16. Jan 16, 2012 #15

    Bacle2

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    My point is that once the expression is fixed, there is a unique, trivial solution, because every element divides every element, by the definition of field.

    Then a|c is tautological for every non-zero element in the field, because every element is a unit, then a|c , inthe sense that a/c :=ac-1 , which is in F by closure properties of F.

    Sorry, my connection is not good tonight.
     
  17. Jan 16, 2012 #16

    Bacle2

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    Anyway, this thread has become an exercise in futility and misunderstanding (misunderestimating?).

    Yes, I guess I could have avoided bringing up the UFD issue, which does not really help, to say the least. My bad.


    Let me delete my posts; they only seem to add confusion. Just wanted to ask.
     
    Last edited: Jan 16, 2012
  18. Jan 16, 2012 #17

    Dick

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    You are right, of course. Not having any irreducible elements does NOT mean a field is not a UFD. Sorry, I seem to have gotten stuck in that impression. May have been too late.
     
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