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

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SUMMARY

This discussion clarifies the analogy between ideals in rings and normal subgroups in groups. An ideal in a ring functions similarly to a normal subgroup by allowing the formation of quotient structures, specifically R/I for a ring R and an ideal I. The key distinction lies in the operations involved; while normal subgroups relate to group operations, ideals involve both additive and multiplicative operations. Understanding this analogy is crucial for grasping the structure of ideals in ring theory.

PREREQUISITES
  • Basic understanding of group theory and normal subgroups
  • Familiarity with ring theory and the concept of ideals
  • Knowledge of ring homomorphisms and their properties
  • Understanding of quotient structures in algebra
NEXT STEPS
  • Study the properties of ring homomorphisms and their kernels
  • Explore the concept of quotient rings, specifically R/I
  • Investigate the relationship between additive and multiplicative operations in rings
  • Learn about the formal definitions of ideals and their applications in ring theory
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Mathematicians, algebra students, and educators seeking to deepen their understanding of the relationship between ideals in rings and normal subgroups in groups.

matheinste
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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.

Matheinste.
 
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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.
 
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.

Matheinste
 
Given a group G and a subgroup H the only way to create G/H with well-defined operations is for H to be a normal subgroup of G.

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

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

-Wolfgang
 
matheinste said:
... an ideal in a ring as 'analogous' to a normal subgroup of a group but there appears to be a slight difference...


yes, and that's because 'analogous' and 'are identical' are not the same thing...
 
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.
 
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.
 
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.

Matheinste.
 

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