Quotient by an ideal versus quotient by extension of the ideal

Question

Let $A \to B$ be a homomorphism of commutative rings. Let $I \subset A$ be an ideal of $A$ and $I^e$ be its extension to $B$. Does the homomorphism $A \to B$ induce a homomorphism $A/I \to B/I^e$? Is it an isomorphism? Are there conditions $A,B,I$ or on the homomorphism $A \to B$ so that they're isomorphic as rings?

Answer 1

Yes, $\varphi\colon A\longrightarrow B$ induces a homomorphism $\;\overline {\mkern-1.5mu\varphi\mkern -1mu}\,\colon A/I\longrightarrow B/IB$, since $I\subset IB$.

It has no reason to be an isomorphism. A simple counter-example is the canonical injection from a domain $A$ which is not a field, into its field of fractions $K$. For any non-zero ideal $I$ of $A$, $IK=K$, so the induced map $A/I$ to $K/IK$ is $0$.

If $\varphi$ is surjective, $\,\overline {\mkern-1.5mu\varphi\mkern -1mu}\,$ is trivially surjective. That's all we can say in general.

However, as $\;\ker\,\overline {\mkern-1.5mu\varphi\mkern -1mu}\,=\varphi^{-1}(IB)/I$, if $B$ is faithfully flat $A$-module, $\,\varphi^{-1}(IB)=I$, so the induced map is an injective ring homorphism.