Ring isomorphism between two quotient rings

Asked Jun 10 '21 at 01:36

Active Jun 10 '21 at 02:00

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Let $f: R \to R'$ be a ring homomorphism where $R$ and $R'$ are two rings, and let $K = \mathrm{ker}f$ be the kernel of the homomorphism. Prove that if $I$ is an ideal of $R$ and $K \subseteq I$, then $R/I \cong R'/f(I)$.

My first thought was to apply transitivity on the ring isomorphism, that is, trying to find a ring $R''$ that connects them together. However, I could not find such a ring. Next, I thought of the First Isomorphism Theorem for Rings to get the proof, but I just realized that that it may be a special case for $R/I \cong R/f(I)$, since $f(\mathrm{ker}f) = {0} \Rightarrow R/\mathrm{ker}f \cong R'$?
Plus, I know if $I$ is an ideal of $R$, then $f(I)$ is an ideal of $R'$, although I'm not sure if this helps.
Thank you in advance!

edited Jun 10 '21 at 02:00

asked Jun 10 '21 at 01:36

Mark Steve

Do you mean $R/I \cong R’/f(I)$? – Sam Freedman Jun 10 '21 at 01:45
There's also a problem that $f$ is not assumed to be surjective, or at least this is not stated. Please double check the hypotheses. – Arturo Magidin Jun 10 '21 at 01:49
Is this what you seek? If so let me know so we can mark it as a dupe. – Bill Dubuque Jun 10 '21 at 02:00
1

The statement is false as written. Take $R'\neq 0$, and $f$ the zero map. If you require rings and morphisms to be unital, take $R=\mathbb{Z}$, $R'=\mathbb{Q}$, $f$ the standard embedding, and $I=\mathbb{Z}$. – Arturo Magidin Jun 10 '21 at 02:09
The image of an ideal need not be an ideal, as Arturo's comment shows. You absolutely need surjectivity here. – paul blart math cop Jun 10 '21 at 02:31
@SamFreedman Yes, this was a typo, thanks for the correction. – Mark Steve Jun 10 '21 at 03:42

Ring isomorphism between two quotient rings

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