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Primes in set of rational numbers

  1. Jan 21, 2010 #1
    There was a part c and d from a question I couldn't answer.
    Let [tex]R = \{ a/b : a, b \in \mathbb{Z}, b \equiv 1 (\mod 2) \}[/tex].

    a) was find the units, b) was show that [tex]R\setminus U(R)[/tex] is a maximal ideal. Both I was successful. But

    c) is find all primes, which I believe i only found one....the rational number 2.

    d) find all ideals and show that [tex]R[/tex] is a PID.

    Any help would be appreciated.
     
  2. jcsd
  3. Jan 22, 2010 #2
    c) Write an element,
    [tex]p = 2^n \frac{a}{b}[/tex]
    where a and b are odd. If n>1, then let, [itex]x = 2^{n-1}[/itex] and [itex]y =2[/itex]. Then [itex]p | xy[/itex], but [itex]p \not | x[/itex] and [itex]p \not| y[/itex]. Thus p is not a prime if n>1.
    If n=1, and, p|xy, then either x or y must have a factor 2. Assume WLOG that 2|x, then p|x.
    If n=0, then p is a unit.
    Thus only associates of 2 are prime.

    d) For some ideal X of R let n be the largest non-negative integer such that [itex]2^n[/itex] divides all elements in X. Then [itex]2^n \in X[/itex] since there exists odd integers a,b such that [itex]2^n \frac{a}{b} \in X[/itex], but then [itex]2^n\frac{a}{b}\frac{b}{a}=2^n[/itex]. We have [itex]X=(2^n)[/itex].
     
  4. Jan 22, 2010 #3
    doesn't 1 = 1 mod 2?
     
  5. Jan 22, 2010 #4
    Yes this is true, but how does it relate to anything in any of the previous posts?
     
  6. Jan 22, 2010 #5
    i didn't read your proof but doesn't my comment imply that for every prime p in Z the element r=p/1 is in R
     
  7. Jan 22, 2010 #6
    I misunderstood your comment. math_grl very likely means prime element in the general sense of prime elements in an integral domain. In an integral domain R we say that a non-unit p is a prime element if p|ab imply p|a or p|b. This is equivalent to the ordinary notion of prime numbers in Z, but it need not be the same for other rings.

    In the R given by math_grl 3 for instance is not a prime because it's a unit (i.e. 1|3 and 3 |1). To see this note that 3 = 3*1 so 1|3 and 1 = 3*1/3 so 3|1. On the other hand 6 is prime in R.
     
  8. Jan 23, 2010 #7

    HallsofIvy

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    But ice109's point is, I believe, that all of the usual primes still are primes in this domain.
     
  9. Jan 23, 2010 #8
    But they aren't. All the usual odd primes in Z are units in this domain and therefore not primes. This is due to the fact that if p is an odd prime in Z, then 1/p is in the domain and since p(1/p)=1 we have p|1. He is of course completely right if prime element is taken to mean prime element in Z, but as we're working in R it would seem natural to consider prime elements in R.
     
  10. Jan 25, 2010 #9
    Last edited by a moderator: Apr 24, 2017
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