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Math Help - Field and ideals question

  1. #1
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    Field and ideals question

    Can someone explain why a field can have no non-trivial ideals?
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  2. #2
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    Re: Field and ideals question

    suppose we have a field F (which is, of course, also a ring). let I be an ideal of F that contains x ≠ 0.

    since x ≠ 0, we have 1/x in F, and since I is an ideal 1 = (1/x)(x) is also in I.

    thus for ANY a in F, a = (a)(1) is in I, so that I = F.

    (in any ring with identity, if an ideal I contains 1, it is the entire ring).
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  3. #3
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    Re: Field and ideals question

    Why is it that since I is an ideal 1 = (1/x)(x) is also in I?
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  4. #4
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    Re: Field and ideals question

    for an ideal,I, of a ring R, we have:

    for any x in I, and all r in R: rx is in I (ideals "absorb" elements when you multiply by them).

    an ideal is an additive subgroup that "acts like 0":

    I + I = I
    -I = I
    RI = IR = I

    just like:

    0 + 0 = 0
    -0 = 0
    a0 = 0a = 0

    it is these "zero-like" properties that allow ideals to serve as "ring-kernels" (when we form a quotient ring R/I, we are essentially "setting all of I equivalent to 0" since I is the additive identity of R/I).
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  5. #5
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    Malavarayanatham
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    Re: Field and ideals question

    Can someone explain why a field can have no non-trivial ideals?
    Ans:
    If R is a field and U is any nontrivial ideal of R then 1 belongs to U then by theorem U = R for any Ideal U of the field R.
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