Is $\ker(B \to A/I \otimes_A B)$ the ideal generated by $f(I)$? Let $A$ and $B$ be commutative rings with units, let $f: A \to B$ be a ring homomorphism, and let $I \subset A$ be an ideal. Then $A/I$ and $B$ are naturally $A$-modules, so we have a tensor product $A/I \otimes_A B$. Let $J$ be the kernel of the surjective map
$$B \to A/I \otimes_A B, \quad b \mapsto 1 \otimes b.$$
It is easy to see that $J$ contains the ideal generated by $f(I)$. Is it, in fact, an equality?
 A: Let $R$ be a commutative ring. Given $R$-linear and surjective maps $\varphi \colon M \to M’$ and $\psi \colon N \to N’$, then, if $i \colon \ker \varphi \to M$ and $j \colon \ker \psi \to N$ are the inclusion maps,
$$
\ker(\varphi \otimes_R \psi) = \text{im}(i \otimes_R \text{id}_N) + \text{im}(\text{id}_M \otimes_R j).
$$
Proof: See theorem $2.19$ of Keith Conrad’s notes: Tensor products II. $\square$

Now, for your problem, note that if $\pi \colon A \to A/I$ is the canonical quotient map, and $\mu \colon A \otimes_A B \to B$ is the isomorphism such that
$$
\mu(a \otimes b) = ab := f(a)b,
$$
then your map makes the following diagram commute.
$$
\require{AMScd}
\begin{CD}
A \otimes_A B @>{\pi \otimes_A \text{id}_B}>> (A/I) \otimes_A B \\
@V{\mu}VV @| \\
B @>>> (A/I) \otimes_A B
\end{CD}
$$
Hence, the kernel of your map is $\mu(\ker(\pi \otimes_A \text{id}_B ))$, which, by the above proposition, is $\mu(\text{im}(i \otimes_A \text{id}_B))$, where $i \colon I \to A$ is the inclusion map.
Finally, since every element of $I \otimes_A B$ (the domain of $i \otimes_A \text{id}_B$) is a sum of the form $\sum_i a_i \otimes b_i$, with $a_i \in I$ and $b_i \in B$, we then have that every element of $\mu(\text{im}(i \otimes_A \text{id}_B))$ is a sum of the form $\sum_i f(a_i)b_i$; so, $\mu(\text{im}(i \otimes_A \text{id}_B))$ is the ideal of $B$ generated by $f(I)$.
