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I am at a loss as to proving the following result:

Let $\phi:R \to S$ be a ring homomorphism with $R$ commutative and $I$ an ideal of $R$ such that $I \subseteq ker \phi$ . Then there exists a ring homomorphism,

$\overline{\phi}: R/I \to S $ such that, $\overline{\phi} \circ \pi = \phi$

Clarifying a few things above, $R/I$ denotes the quotient ring of $R$ by $I$ and $\pi$ denotes the canonical homomorphism,

$\pi : R \to R/I$ with $\pi(r)=r + I$ for $r \in R$

It may help to think of this in terms of a triangular commutative diagram. Any assistance would be appreciated, thanks :-)

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The assumption that $R$ is commutative is not necessary.

The only possible definition of $\bar\phi$ is $$ \bar\phi(r+I)=\phi(r) $$ and you just need to show this is a good definition; that is, if $r+I=r'+I$, then $\phi(r)=\phi(r')$.

But $r+I=r'+I$ means $r-r'\in I$, so, by assumption, $r-r'\in\ker\phi$, so…

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  • $\begingroup$ Ahh right - but what is the step in logic to get $\phi(r')=\overline{\phi}(r+I)$ ? $\endgroup$ – S Valera May 14 '14 at 21:54
  • $\begingroup$ Actually, no that's stupid, I get it! Thanks a million. $\endgroup$ – S Valera May 14 '14 at 21:56
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You may want to look at this and do some diagram chasing: $$\begin{matrix}0\to&\ker\phi&\to& R&\stackrel\phi\to &S\\ &\;\uparrow\rlap{\scriptsize\subseteq}&&\|&&\;\uparrow\rlap{\scriptsize\bar\phi}\\ 0\to &I&\to& R&\stackrel\pi\to& R/I&\to& 0 \end{matrix} $$

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  • $\begingroup$ This helped shed some light on the inner-workings of the result via the structure. A useful contribution, thank you. $\endgroup$ – S Valera May 14 '14 at 22:10

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