Modify this formula : $R/I \cong \phi[R]/\phi[I]$

Question

Let $R$ be a ring and $I$ an ideal of $R$, and let $\phi : R\longrightarrow R'$ be a ring homomorphism.

Studying by myself, I have a conjecture the following: $$R/I \cong \phi[R]/\phi[I].$$ This formula is such a beautiful result. But, I found the error when the homomorphism is not one-to-one. So, I want to modify this formula as possible.

My trial is the following. First, assume the homomorphism $\phi$ is one-to-one. But, this revision may be useless because its assumption is too strong. Second, in order to use the first isomorphism theorem, if $\psi : R\longrightarrow \phi[R]/\phi[I]$ is a canonical homomorphism, then the modification of our formula is $R/\ker\psi \cong \phi[R]/\phi[I]$.

Except the above modifications, do you have another modification? Please let me know!

In the case that $\phi$ is injective, everything is trivial, since $\phi$ becomes an isomorphism onto $\phi (R)$. — Crostul, Aug 23 '15 at 13:19

user26857 · Accepted Answer · 2015-08-23T22:22:03.690

3

In fact, $$\phi(R)/\phi(I)\simeq R/(I+\ker\phi)$$ and this follows easily by the fundamental theorem of isomorphism for rings. (Find the kernel of the surjective homomorphism $R\to\phi(R)\to\phi(R)/\phi(I)$.)

edited Aug 23 '15 at 22:22

answered Aug 23 '15 at 20:44

user26857

52,094

Is it right the surjective homomorphism you wrote is the very $/psi$? that is, $\psi(a)=\phi(a)+\phi[I]$ for each $a\in R$. – JeongHobin Aug 23 '15 at 21:42
Ah, I see. I had a subtle problem :) Thanks! – JeongHobin Aug 23 '15 at 22:20

Modify this formula : $R/I \cong \phi[R]/\phi[I]$

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