Is Every Ideal in a Ring the Kernel of a Homomorphism?

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SUMMARY

In the discussion, it is established that while all kernels of ring homomorphisms are indeed ideals, the converse is not universally true. Specifically, for any ideal I of a ring R, there exists a homomorphism f: R -> R/I such that the kernel of f is precisely I. This mapping is confirmed by the example provided, where elements of R are mapped to their corresponding cosets in R/I.

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  • Understanding of ring theory and ideals
  • Familiarity with ring homomorphisms
  • Knowledge of quotient rings, specifically R/I
  • Basic concepts of algebraic structures
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Mathematicians, algebra students, and anyone studying abstract algebra, particularly those focused on ring theory and homomorphisms.

joecoz88
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Kernel <--> Ideal?

I know that all kernels of ring homomorphisms are ideals, but is it true that for any ideal I of a ring R, there exists a homomorphism f: R -> R' such that Ker(f)=I?
 
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You map R to R/I by x->xI
 


yep!

as Shredder says!
 

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