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Number theory ideals proof, where am I going wrong?

  1. Feb 22, 2012 #1
    21aa0pc.png

    I'm trying to prove part iii)

    So far:

    Show x irreducible => no y in D-Dx where <x> is a proper subset of <y>

    Suppose the contrary that x is reducible

    => x = ay for some a,y in D-Dx

    => x is an element of <y>

    => <x> is a subset of <y>

    By part i) we showed that if <x>=<y> then a must be a unit, but this contradicts the fact that a is in D-Dx.

    => <x> must be a proper subset of <y>

    Here is where I'm stuck

    How do I complete this, or have I just gone the wrong way about it? I think I may be proving the converse of the iff statement in reverse, instead of proving this part of it.

    Thanks!
     
  2. jcsd
  3. Feb 22, 2012 #2

    jbunniii

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    If <x> is a proper subset of <y>, then <x> can't be all of D, right?
     
  4. Feb 22, 2012 #3
    So this tells me that x cannot be a unit and then what? :)
     
  5. Feb 22, 2012 #4

    jbunniii

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    No, follow your train of logic. You have proven this:

    x reducible => <x> is a proper subset of D

    which is logically equivalent to this:

    <x> = D => x is not reducible
     
    Last edited: Feb 22, 2012
  6. Feb 22, 2012 #5
    Hmm hang on, for the whole thing wouldnt it be simpler for me to suppose x is irrecudible

    => x = ay for some unit a, y is in D-Dx else x=1

    => x is in <y>

    => <x> is contained in <y>

    by part i) then <x> =<y> since x and y are associates.

    So there can be no y in D-Dx such that <x> is a proper subset of <y>

    Then i just prove in the other direction! No?
     
  7. Feb 22, 2012 #6

    jbunniii

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    I'm not sure why you are focusing on reducible vs. irreducible. (ii) asks you to prove that <x> = D if and only if x is a unit. So why not start by assuming that x is a unit, and showing that this implies <x> = D?
     
  8. Feb 22, 2012 #7
    I'm trying to prove part iii)! :)
     
  9. Feb 22, 2012 #8

    jbunniii

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    LOL, sorry! I misread your original post. I'll go back and read it again with this in mind.
     
  10. Feb 22, 2012 #9

    jbunniii

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    Suppose x is irreducible, and suppose x = ay for some elements a and y of D. If y is not a unit, then a must be a unit since x is irreducible.

    Therefore, [itex]y = a^{-1}x[/itex], so [itex]<x> \subset <y>[/itex] and [itex]<y> \subset <x>[/itex]. Therefore <x> = <y>.

    Now, for the other direction, you can do it one of two ways.

    1. Assume x is NOT irreducible. Thus x is either a unit or a product of two non-units. Consider both cases and try to prove the conclusion.
    or
    2. Assume that there IS a nonunit y such that <x> is properly contained in <y>. Prove that x is irreducible.
     
    Last edited: Feb 22, 2012
  11. Feb 22, 2012 #10
    Ok, I'm using option 2 here

    Assume that there IS a nonunit y such that <x> is properly contained in <y>

    So by part i) we have y divides x where x&y are not associate

    => x=ay where a is non unit

    => x is reducible

    Then my arguement is 'so there are no cases where y is a non unit such that x is irreducible, so x can only be irreducible if y is a unit'

    Is that ok? I'm having a hard time believing I've proved it all

    What I see is that I proved if there IS a nonunit y such that <x> is properly contained in <y> then x is reducible, and that if y is a non unit then x may be irreducible, and not what I set out to prove!
     
  12. Feb 22, 2012 #11

    jbunniii

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    Yes, your proof looks OK to me. Let's recap what you have proved.

    "if there IS a nonunit y such that <x> is properly contained in <y> then x is reducible"

    This is logically equivalent to:

    "if x is irreducible, then there is NO nonunit y such that <x> is properly contained in <y>"

    which is precisely the => direction of the statement you are trying to prove.

    In post #9, I also proved the => direction, using a different approach.

    But you still need to prove the <= direction.

    I listed two options to prove the <= direction, but unfortunately I made a typo in my option 2! Not enough coffee this morning, I guess.

    It should have been:

    1. Assume x is NOT irreducible. Thus x is either a unit or a product of two non-units. Consider both cases and try to prove the conclusion.
    or
    2. Assume that there is NO nonunit y such that <x> is properly contained in <y>. Prove that x is irreducible.

    You took my original option 2 ("Assume that there IS a nonunit y such that <x> is properly contained in <y>. Prove that x is irreducible." - a false statement) and turned it into "Assume that there IS a nonunit y such that <x> is properly contained in <y>. Prove that x is reducible" - a true statement, but equivalent to the => direction!
     
  13. Feb 22, 2012 #12
    Ah, so I still need to prove the <= direction?

    So using 1) this time

    I assume this is logically equivelent to

    The negation of (so I need to prove this following statement is not the case): [x reducible => there is a y non unit (where <x> is a proper subset of <y>)]

    Which is equal to: [there is no such y non unit (where <x> is a proper subset of <y>) => x irreducible]

    I'm taking this from what I believe:

    -(A=>B) = (-B=>-A)

    Is this true, and is this what you want me to do?
     
  14. Feb 22, 2012 #13
    ok perhaps I'm complicating it, if I just do number 2)

    Assume that there is NO nonunit y such that <x> is properly contained in <y>. Prove that x is irreducible.

    if thre were such y then by part i) we have y divides x where x&y are not associate

    But no such y exists in our assumption so x&y have to be associate, which means x=ay for all y, so x is irreducible.

    Is that ok?

    Sorry the fact that there is NO nonunit y is very confusing to me, making this proof very confusing!
     
  15. Feb 22, 2012 #14

    jbunniii

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    Yes, that's correct. Your negation "[there is no such y non unit (where <x> is a proper subset of <y>) => x irreducible]" is the same as my revised option 2:

    2. Assume that there is NO nonunit y such that <x> is properly contained in <y>. Prove that x is irreducible.
     
  16. Feb 22, 2012 #15

    jbunniii

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    I think you're running in circles. If the thing you are supposed to prove starts with "Assume that there is NO nonunit y such that..." then your proof can't start with "If there were such y..."

    I think option 1 will be easier to work with:

    1. Assume x is NOT irreducible. Thus x is either a unit or a product of two non-units. Consider both cases and try to prove the conclusion.

    i.e. If x is a unit or a product of non-units, then there is a nonunit y such that <x> is properly contained in <y>.

    I suggest considering the two cases separately: first assume x is a unit, and show that there is a nonunit y such that <x> is properly contained in <y>.

    Then assume x is a product of two nonunits, and show the same conclusion.
     
  17. Feb 22, 2012 #16
    I thought by post #12 we are trying to show the negation of (so disprove)

    'If x is a unit or a product of non-units, then there is a nonunit y such that <x> is properly contained in <y>.'

    but you're asking me to prove it!

    Also as a side note, why do I have to prove if x is a unit? If x is a unit then it's irreducible isn't it?
     
  18. Feb 22, 2012 #17

    jbunniii

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    This is the second half of the "if and only if"!

    You are trying to show that "A if and only if B."

    You have already proven "if A, then B."

    Now you want to prove "if B, then A", or equivalently, "if not A, then not B."

    This statement: 'If x is a unit or a product of non-units, then there is a nonunit y such that <x> is properly contained in <y>.' is the "if not A, then not B" I am referring to.

    No, by definition an irreducible element is NOT a unit. This is easiest to remember if you think of the special case of the integers. The units are +/-1, and the irreducibles are +/-p where p is any prime number.
     
  19. Feb 22, 2012 #18
    Ohh I see now!

    I can do it now I'm sure!

    But last question about the unit case

    If x is a unit, then the ideal generated by x is the domain D, which is certainly not a proper subset of whatever the ideal generated by y is. Are you sure I need this case?
     
  20. Feb 22, 2012 #19

    jbunniii

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    Yes, I just noticed this too. I think the statement is actually not correct as stated.

    I think (iii) should read:

    "x is irreducible OR A UNIT if and only if there is no nonunit y such that <x> is properly contained in <y>"

    Maybe your instructor or book is allowing units to be irreducible? I don't think that is a standard definition, but if that is their definition then the statement would be valid.
     
  21. Feb 22, 2012 #20

    Deveno

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    yes. only non-units are irreducible.

    i think that (iii) should be:

    a non-unit x is irreducible iff....

    because a unit certainly has the property that there is no non-unit y (indeed no unit nor non-unit y) such that <x> is a proper subset of <y>, but units are not generally considered irreducible.
     
    Last edited: Feb 22, 2012
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