1. Not finding help here? Sign up for a free 30min 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!

Showing the quotient is not a UFD

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that [itex] \mathbb Q[x,y]/(x^2+y^2-1) [/itex] is not a unique factorization domain.

    3. The attempt at a solution

    We have tried a few approaches. Using [] to denote equivalence classes, we note that we can write [itex] [x]^2 = [1-y][1+y][/itex]. Our goal was to show that this is a non-unique prime decomposition, but this doesn't work since neither of these elements are prime. Similarly, [itex] [x+y-1][x+y+1] = [2x][y] [/itex] but the same problem applies.

    Next we tried playing around with evaluations of polynomials. In particular, in the quotient ring we can write any element as [itex] p(x) + y q(x) [/itex]. Taking a product of such polynomials and evaluating at zero gives a structure similar to [itex] \mathbb C[/itex].

    Another suggestion was to use Laurent polynomial. Any ideas?
     
  2. jcsd
  3. Nov 28, 2011 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Does your definition of UFD require a unique decomposition into primes or a unique decomposition into irreducibles?
     
  4. Nov 28, 2011 #3
    Both. They are equivalent in a UFD no?
     
  5. Nov 28, 2011 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If it's not a UFD then it's not the same thing (namely, elements can be irreducible without being prime). So your first attempt at x2=(1-y)(1+y) is a good place to look... it's not a prime factorization, but they are irreducible factorizations
     
  6. Nov 28, 2011 #5
    Yeah, that had occurred to me, though I'm not too sure how to show that [itex] 1 \pm y [/itex] is irreducible. It seems obvious since they're monic first order polynomials, but I don't think I can just claim that (can I?). In any case, I would like to use Eisenstein's criterion but then I run into the problem of dealing with the prime ideals of [itex] \mathbb Q [/itex], of which there are no non-trivial proper ideals. Should I perhaps be looking at the prime ideals of in [itex] \mathbb Q[x] [/itex] instead?
     
  7. Nov 28, 2011 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I fear you need to start with the very definition of irreducible. You ring is pretty strange, so you can't use Eisenstein.

    Assume that y+1=b*c and obtain a contradiction.
     
  8. Nov 28, 2011 #7
    I think what you're hinting at is a degree argument, in that the only way that product could be a first order polynomial is precisely when one is a constant and the other is a first order polynomial. Since [itex] \mathbb Q[/itex] is a field, the constant is a unit, so we can conclude that 1-y is irreducible. However, does this necessarily apply in the quotient? Degree of equivalence classes is definitely not preserved in the quotient so I don't see how that argument would work.

    Edit: The degree of the product of equivalence classes of polynomials in the quotient is not well behaved, is what I meant to say. That's a mouthful!
     
  9. Nov 28, 2011 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    A direct attack would indeed not work. But let's look at it this way:

    Assume

    [tex][y+1]=[P(x,y)][Q(x,y)][/tex]

    Then

    [tex]y+1-P(x,y)Q(x,y)\in (x^2+y^2-1)[/tex]

    So

    [tex]y+1-P(x,y)Q(x,y)=R(x,y)(x^2+y^2-1)[/tex]

    and now try to obtain a contradiction.
     
  10. Nov 28, 2011 #9
    Sorry. I've been trying to play around with this but I don't see it.

    Okay. So I don't think this is a degree argument. I say this because there is no restriction on the degree of x in this case. The degree of y could have lead somewhere, but the presence of the -1 really means that y can take on any degree.

    So I'm guessing this is a divisibility argument. I tried arguing that by re-arranging, we could get
    [tex] (y+1)(R(x,y)(y-1) - 1) = -P(x,y)Q(x,y) - R x^2 [/tex]
    in which case since y+1 divides the left hand side, it must also divide the right hand side. Though I don't see anything coming from this.
     
  11. Nov 29, 2011 #10
    If I could show that [x] is irreducible and that the only the image of constants in [itex] \mathbb Q[x,y]/(x^2+y^2-1) [/itex] are units, could I show that it is not a UFD?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook