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

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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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