2
$\begingroup$

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 :-)

$\endgroup$
3
$\begingroup$

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…

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

You may want to look at this and do some diagram chasing: $$\begin{matrix}0\to&\ker\phi&\to& R&\stackrel\phi\to &S\\ &\;\uparrow\rlap{\scriptsize\subseteq}&&\|&&\;\uparrow\rlap{\scriptsize\bar\phi}\\ 0\to &I&\to& R&\stackrel\pi\to& R/I&\to& 0 \end{matrix} $$

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.