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Proving an ideal is not principal, have I gone wrong?

  1. May 8, 2010 #1
    Prove (3, 2 + √−5)R is not principal in Z[√−5].

    Solution, where N = norm:

    3 = m(a+b√−5) => N(a+b√−5) | N(3) = 9

    and

    2+√−5 = n(a+b√−5) => N(a+b√−5) | N(2+√−5) = 9

    n,m in R

    So,

    N(a+b√−5) = a2+5b2 = 1, 3 or 9

    3 cannot be a solution clearly.

    If N(a+b√−5) = 9 then => a=+-3 b=0 , or a=+-2 b=+-1

    Take ideal I = (3)R

    => 2+√−5 = 3r, r is in R

    => N(2+√−5) = 9.N(r)
    => 9 = 9.N(r)

    so r is a unit, so all of r must be units in R => R is a field => only ideals in R are {0R} and R which isn't (3)R.

    For a=2 b=1

    => (2+√−5)r = 3, r is in R, so by same arguement above this cannot be an ideal.

    For N(a+b√−5) = 1

    We have a=+-1 b = 0

    Take I = (1)R = R

    => 1 = 3m + (2+√−5)n

    => (2-√−5) = 3(2-√−5)m + 9n

    => 3 | (2-√−5)

    => 3k = 2-√−5

    => 9.N(k) = 9 => k unit and by above can be an ideal in R. But R is clearly not a field so this cannot be an ideal?

    How do I show this is not principal?

    Is the rest of this ok?

    Wrong techniques?
     
    Last edited: May 8, 2010
  2. jcsd
  3. May 8, 2010 #2

    Office_Shredder

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    This doesn't make sense. Just because r is a unit doesn't mean every element of R is a unit.
     
  4. May 8, 2010 #3
    well it wasnt in my initial solution, as i was typing it it just occured to me so i'd thought i'd throw it in there

    how can i show it's not principal then? :/
     
  5. May 8, 2010 #4
    I don't see how it's not correct though, the condition was to show

    2+√−5 = 3r, r is in R for all r

    surely? Since we're trying to see if 3 is a generator, and the only way r can fit here is if it's a unit, so this contradicts the fact that Z[−5] is not a field.

    or maybe because Z[√−5] is not a Unique factorization domain so when r is a unit, r = +-1, then this cannot hold?
     
  6. May 8, 2010 #5

    Office_Shredder

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    You need that 2+√−5 = 3r is true for ONE VALUE of r in R, not for every possible r in R. So all this demonstrates is that R has a non-trivial unit, not that R has only units.

    You can check very directly that no such r exists. If [tex]r=a+b\sqrt{-5}[/tex] then [tex]2+\sqrt{-5}=3a+3b\sqrt{-5}[/tex] And obviously no values of a and b work here
     
  7. May 8, 2010 #6
    Here's a sketch of how I'd prove the result:

    First, if a, b [itex]\in[/itex] R, we'll let a~b mean that a and b are associates in R (that is, a = ub, where u is a unit in R).

    First, show that both 3 and (2 +[itex] \sqrt{-5}[/itex]) are irreducible in R.

    Then suppose that [itex](3, 2 + \sqrt{-5})_R}[/itex] = [itex](\alpha)_R[/itex], with [itex]\alpha \in [/itex] R. This implies that [itex]\alpha[/itex]|3, so either [itex]\alpha[/itex]~1 or [itex]\alpha[/itex]~3 (since 3 is irreducible in R). If [itex]\alpha[/itex]~3, then 3|(2 +[itex]\sqrt{-5}[/itex]). Show that this is impossible. Thus, [itex]\alpha[/itex]~1, so [itex](\alpha)_R[/itex] = R.

    Therefore, 1 [itex]\in (\alpha)_R[/itex]. Assume that 1 = 3(a + b[itex]\sqrt{-5}[/itex]) + (c + d[itex]\sqrt{-5}[/itex])(2 + [itex]\sqrt{-5}[/itex]) and derive a contradiction.

    Petek
     
  8. May 8, 2010 #7
    doesnt this require R to be a UFD?
     
  9. May 8, 2010 #8
    I think that all we need is the the ring be commutative with an identity element and (possibly) that it be an integral domain for this definition to make sense. However, if you haven't run into this concept, then my solution might not make sense.

    Petek
     
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