Prove that $R[x]/(I,f)$ is isomorphic to $(R/I)[x]/(f')$, where $f'$ is $f$ in $R/I$. Prove that $R[x]/(I,f)$ is isomorphic to $(R/I)[x]/(f')$, where $f'$ is $f$ in $R/I$.
Attempt: I know how to prove $R[x]/(I) \cong (R/I)[x]$ using the first isomorphism theorem and the homomorphism that sends $f \in R[x]$ to $f' \in (R/I)[x]$.
Question: May I have a hint on how to proceed next?
 A: I assume that $R$ is a commutative ring, $I$ is an ideal of $R$ and $f \in R[x]$? Recall the universal properties:


*

*If $S$ is a commutative ring and $s \in S$ and $R \to S$ is a homomorphism, there is unique homomorphism $R[x] \to S$ which extends $R \to S$ and maps $x$ to $s$.

*If $S$ is a commutative ring and $R \to S$ is a homomorphism which vanishes on $I$, then it extends uniquely to a homomorphism $R/I \to S$.


Using these, it is easy to construct homomorphisms in both directions, which are then (by the uniqueness property applied to the compositions) inverse to each other. For example, we have $R \to R/I \to R/I [x] \to R/I [x]/(\overline{f})$ and the (class of the) element $x$ in $R/I [x]/(\overline{f})$, hence a homomorphism $R[x] \to R/I [x]/(\overline{f})$, which clearly vanishes on $f$ and on $I$, hence induces a homomorphism $R[x]/(I,f) \to R/I [x]/(\overline{f})$. The homomorphism in the other direction is constructed in the same way.
If you already know the Yoneda Lemma, here is a fast proof:
$\hom(R[x]/(I,f),S)\\\cong \{(\alpha,s) : \alpha \in \hom(R,S), s \in S, \alpha|_I = 0, \alpha(f)(s)=0\} \\\cong \{(\beta,s) : \beta \in \hom(R/I,S), s \in S, \beta(\overline{f})(s)=0\} \\\cong \{\gamma \in \hom(R/I [x],S) : \gamma(\overline{f})=0\}\\ \cong \hom(R/I [x]/(\overline{f}),S)$
