When $B$-modules $M$ and $M_A \otimes_A B$ are isomorphic?

Question

Let $\varphi:A \to B$ a ring map and $M$ a (right)-$B$-module. We can associate to $M$ by restriction along $\varphi$ the (right)-$A$ module $M_A$, where the $A$-multiplication is given by $m \cdot a := m \cdot \varphi(a)$. Let $N:= M_A \otimes_A B$. $N$ is by construction an $B$ module.

Is there a sufficient & neccessary characterization on $A$ and $B$ known when $M =N$ as (right)-$B$-modules?

If $\varphi:A \to B:=A/I \cdot A$ is quotient for some ideal $I$ of $A$, then the $M_A$ 'not forgets' the $B$-module structure and then $M=N$ holds. Is this the only case?

I expect that this is a neccessary and sufficient condition, because it feels that if $\varphi$ is not surjective, $M_A \otimes_A B$ 'forgets' the action on $M$ by elements in $B$ not in the image. But I not know how to prove it.

Answer 1

If the question you're asking is when the natural map $M \otimes_A B \to M$ is an isomorphism (note that this is slightly stronger than just asking that the two be abstractly isomorphic), the answer is that this happens for all $M$ iff it happens in the special case that $M = B$ (since both the source and the target, as a functor of $M$, preserve colimits), meaning iff the multiplication map

$$B \otimes_A B \to B$$

is an isomorphism.

If $A$ and $B$ are both commutative, then it's an exercise to check that this condition is equivalent to the map $f : A \to B$ being an epimorphism in the category of commutative rings. This happens, for example, if $f$ is surjective, but also if $f$ is a localization, e.g. the map $f : \mathbb{Z} \to \mathbb{Q}$. I don't know the precise characterization in general (but see this MO question about epimorphisms of commutative rings); in the special case that $A = \mathbb{Z}$ rings $B$ with this property are called solid rings and they were classified by Bousfield and Kan.

I don't know what happens in the noncommutative case. An easy special case is that if $A = k$ is a field and $B$ is finite-dimensional over $k$ then by a simple dimension count the multiplication map $B \otimes_k B \to B$ can't be an isomorphism unless $\dim_k B = 1$, meaning that $B = k$. Maybe one can say something more generally about the case that $B$ is flat over $A$, I don't know.