Base change of injective ring homomorphism

Question

Let $f:A \to B$ be an injective ring homomorphism of $\mathbb{C}$-algebras and $A, B$ are integral domains. Suppose that $B$ is the integral closure of the ring $A$ (in the fraction field of $A$) i.e., $\mathrm{Spec}(B)$ is the normalization of $\mathrm{Spec}(A)$. Let $I$ be an ideal in $A$. Is the induced morphism from $A/I$ to $B \otimes_A A/I$ injective?

Answer 1

$\newcommand{\C}{\mathbb C}$No. Let $A=\C[t^2,t^3]$, so the fraction field is $\C(t)$ and $B = \C[t]$. Take the ideal $I = \langle t^2\rangle$, so $A/I$ has a $\C$-basis $\{1,t^3\}$. In particular, $t^3$ is not a multiple of $t^2$ in $A$, so $t^3$ is not $0$ in $A/I$. However, the image of $t^3$ is $0$ in $B\otimes_A A/I$: $$ A/I \ni t^3\mapsto 1\otimes t^3 = t\otimes t^2 = t\otimes 0 = 0\in B\otimes_A A/I. $$ I guess this would always work unless $A=B$. Take $b\in B\setminus A$. Write $b$ as $a_2/a_1$, where $a_1,a_2\in A$. Then take $I = \langle a_1\rangle$. Since $b\notin A$, $a_2$ is not a multiple of $a_1$ in $A$, so $a_2\notin I$, so $a_2$ is not $0$ in $A/I$. However, its image in $B\otimes_A A/I$ is $0$: $$ 1\otimes a_2 = 1\otimes \frac{a_2}{a_1}a_1 = 1\otimes ba_1 =b\otimes a_1 = 0. $$