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An ideal in a ring as 'analogous' to a normal subgroup of a group, but

  1. Jul 27, 2007 #1
    Hello all. I am in need of a quick clarification.

    A text I am reading describes an ideal in a ring as 'analogous' to a normal subgroup of a group but there appears to be a slight difference in structure in that a member of the underlying additive group from which the ideal is formed operates on a member of the ideal to produce a member of the ideal, at least that is how I read it. Am I mistaken.

  2. jcsd
  3. Jul 27, 2007 #2


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    What do you mean by "operates?" Obviously a normal subgroup and an ideal aren't going to be exactly the same. The analogy is that ideals are kernels of ring homomorphisms and thus can be used to obtain a quotient structure of the ring, just like how normal subgroups are kernels of group homomorphisms and give rise to quotient groups.
  4. Jul 27, 2007 #3
    Thankyou morphism.

    Sorry, bad terminology. I meant the binary operation ( multiplication )between two members of the ring/ideal.

    I'll look up the points you have made. I think I was overlooking the point that the binary operation I was referring to is the 'multiplicative' second operation and not the primary operation of the additive group.

    I am looking really for an informal pointer to the structure of an ideal and then I will be able to understand the formal definition.

  5. Jul 27, 2007 #4
    Given a group [tex]G[/tex] and a subgroup [tex]H[/tex] the only way to create [tex]G/H[/tex] with well-defined operations is for [tex]H[/tex] to be a normal subgroup of [tex]G[/tex].

    In ring theory we have a similar situation. Given a ring [tex]R[/tex] and a subring [tex]N[/tex] to create well-defined operations for [tex]R/N[/tex] we require that [tex]N[/tex] be an ideal of [tex]R[/tex].

    So it is as if it plays the role of the normal subgroup in ring theory.

  6. Jul 28, 2007 #5

    matt grime

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    yes, and that's because 'analogous' and 'are identical' are not the same thing....
  7. Jul 28, 2007 #6
    Thanks Kummer.

    The parallel between Subgroups and Ideals that you have pointed out is likely to be most helpful. I must spend a couple of hours going back to basics. I must learn to walk before I can run but I think the general idea is coming through.

    Thanks Matheinste.
  8. Jul 30, 2007 #7


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    to clarify kummers post further, if H is a subgrop of G, then there is a group operation on G/H = equivalence classes of elements of g under the relation xh is equivalent to x for all h in H, such that the natural map G-->G/H taking x to its equivalence class, is a homomorphism, if and only if H is a normal subgroup.

    similarly, if I is an additive subgroup of the ring R, then the group R/I has a ring structure such that R-->R/I is a ring map, if and only if I is an ideal in R.

    i hope this is right. try proving them.
  9. Aug 1, 2007 #8
    Thankyou all for your comments. I now understand the structure of an ideal but there is still much more additional stuff to take in and I look forward to future help.

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