1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    Bacle2

    User Avatar
    Science Advisor

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: When the Product of Non-Squares is a Square
Loading...