Kernel of a morphism between projective modules and base change

Question

Let $A$ be a ring (Noetherian if it helps) and $f : M \to N$ be a morphism between finite projective $A$-modules.

Let $B$ be an $A$-algebra. Is it true that if $f\otimes_A B$ is injective then $\ker(f) \otimes_A B = 0$ ?

This is obviously not true if one removes the assumption that both $M$ and $N$ are finite projective. For example take $A = \mathbb{Z}$, $B = \mathbb{Z}/2\mathbb{Z}$, $M = \mathbb{Z}$ and $N = \mathbb{Z}/2\mathbb{Z}$ and $f : \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ to be the natural projection.

If it is not true are there properties of $A$ that would ensure the result to hold ?

Answer 1

Let $k$ be a field, $A=M=N=k[X,Y]/(XY)$, $f=$ multiplication by $X$, $B=A/(Y)=k[X]$, so that $\ker(f)=(Y)$, $\ker(f) \otimes_A B =(Y)/(Y^2)$.