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Homework Help: 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


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    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


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




    [tex]y+1-P(x,y)Q(x,y)\in (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?
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