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Let R be a commutative ring and let I be an ideal in R.

  1. Nov 12, 2005 #1
    Let R be a commutative ring and let I be an ideal in R.
    1. If every ideal in R is principal, then every ideal in R/I is principal.
    2. If every ideal in R/I is principal, then every ideal in R is principal.
    I must prove or disprove if either is true or false. Can someone tell me whether either is true or false so I can know how to proceed?
     
  2. jcsd
  3. Nov 13, 2005 #2

    Hurkyl

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    I'm sure someone can, but it isn't good for you.

    This is exactly analogous to someone doing arithmetic problems and asking for an answer key before they try to solve them.

    You shouldn't need to know the answer to a problem before you can solve it. :tongue2:
     
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3
    2 doesn't seem to be true. Consider R=Z[x] and I=2Z[x]+xZ[x]. R/I={0,1}. Every ideal in R/I is certainly principal. However, not ever ideal in Z[x] is principal: for instance, let J={polynomials whose constant terms are even integers}. To show this, suppose J is principal, then J consists of multiples of some element p(x). Since 2 is in J, 2 must be a multiple of p(x), which implies that p(x) must be of degree 0. Since p(x) is in J, it must be +/-2. However, x is in J because it has an even constant. Therefore, x must be a multiple of +/-2. This is impossible since all polynomials involved have integer coefficients. Therefore, J is not principal.

    1 is intuitively true, but so far, no luck in proving it. If every ideal I in R is principal, let I={rc:r in R}. R/I={[a]_c : a in R}. If J is an ideal in R/I, then it must be closed under multiplication. I think I can create a contradiction here assuming J is not principal, but I haven't been able to.
     
  5. Nov 13, 2005 #4

    Hurkyl

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    Your approach for 2 looks good.

    For 1, how far did you get?
     
  6. Nov 13, 2005 #5
    Not very far. I can't seem to use the fact that I is a principal ideal anywhere in my proof. What makes R/I where I is principal so special?
     
  7. Nov 13, 2005 #6

    matt grime

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    Ideals in R/I are lift to ideals in R that contain I
     
  8. Nov 13, 2005 #7
    I'm not entirely sure what you mean by "lift". Do you mean ideals in R/I are contained in I?
     
  9. Nov 13, 2005 #8

    Hurkyl

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    It doesn't.
     
  10. Nov 14, 2005 #9

    CarlB

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    It seems that a way of understanding the relationships between ideals and principal ideals is to look at matrix groups over the reals with matrix arithmetic defining addition and multiplication.

    Some principal ideals for 4x4 matrices:

    [tex]\left(\begin{array}{cccc}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)[/tex]


    [tex]\left(\begin{array}{cccc}.5&.5&0&0\\.5&.5&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)[/tex]

    A non principal ideal:

    [tex]\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)[/tex]

    Carl
     
  11. Nov 19, 2005 #10
    I have the same problem and I still haven't solved it. Any suggestions?
     
  12. Nov 20, 2005 #11

    CarlB

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    On principal ideals.

    Perhaps it is worthwhile recalling the various ways that a primitive ideal can be distinguished from an ideal.

    What I recall is that P is primitive if P cannot be written as the sum of two non trivial ideals. A "trivial" ideal would be zero.

    Carl
     
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