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 :-)