How does extension of restriction of $M$ relate to $M$? Let $A,B$ be rings, $f:B\to A$ be a ring homomorphism, and $M$ be an $A$-module.  We can view $M$ as a $B$-module via restriction, and we may then extend the restriction of $M$ to an $A$-module by considering $A\otimes_B M$.  My question concerns how $A\otimes_B M$ relates to $M$.  
It seems that there is a surjective $A$-module homomorphism $A\otimes_B M\to M$ but that this homomorphism is not always injective.  I would like to understand the kernel of this morphism.  In particular, I'm looking for general conditions where the kernel is equation to the $A$-torsion of $A\otimes_B M$.  
Are there well known situations where this occurs?  I have no problem assuming properties such as $A,B$ being Noetherian integral domains, $M$ being finitely generated, etc.  I would also appreciate a simple example where $M$ does not have torsion yet $A\otimes_B M$ does.
 A: There's a natural short exact sequence of $(A, A)$-bimodules
$$0 \to I \to A \otimes_B A \xrightarrow{m} A \to 0$$
where $m$ is the multiplication map and $I$ is its kernel. Geometrically (in the case that everything is commutative), $I$ is the ideal cutting out the relative diagonal $\Delta : \text{Spec } A \to \text{Spec } A \times_{\text{Spec } B} \text{Spec } A$. 
Tensoring this short exact sequence with an arbitrary left $A$-module $M$, and using the isomorphism $A \otimes_B A \otimes_A M \cong A \otimes_B M$, produces a short exact sequence of left $A$-modules
$$0 \to I \otimes_A M \to A \otimes_B M \to M \to 0$$
since $\text{Tor}_1(A, M) = 0$. So the kernel is precisely $I \otimes_A M$. This always vanishes iff the multiplication map $m$ is an isomorphism, which occurs for example if (let me restrict to the commutative case for safety) $A$ is a quotient or localization of $B$. More generally, if $A$ and $B$ are commutative, the multiplication map $m$ being an isomorphism is also equivalent to the map $f : B \to A$ being an epimorphism of commutative rings. 
There are a lot of other things to say here but I'm not sure what you're interested in. You can think of the map $A \otimes_B M \to M$ as the counit of the natural adjunction between $A$-modules and $B$-modules given by restriction and extension of scalars. As such it is part of the structure of an induced comonad on $A$-modules which can be used to study the question of when an $A$-module descends to a $B$-module. In this setting the multiplication map $m$ being an isomorphism is equivalent to the counit being an isomorphism, which is equivalent to the right adjoint (restriction of scalars) being fully faithful. 
A: Let's assume that $A,B$ are integral domains and $M$ is finitely generated.  Let $K$ be the kernel of the map $A\otimes_B M\to M$.  If $M$ is torsion free as an $A$-module, we always have that $K$ contains the $A$-torsion of $A\otimes_B M$.  For the reverse inclusion let $\mathbb{k}$ be the field of fractions of $A$.  Tensoring the SES $$0\to K \to A\otimes_B M \to M \to 0$$ with $\mathbb{k}$ yields the SES $$0\to \mathbb{k}\otimes_A K\to \mathbb{k}\otimes_B M\to \mathbb{k}\otimes_A M\to 0$$ since $\mathbb{k}$ is a flat $A$-module.  Now by rank nullity, $\mathbb{k}\otimes_A K$ is zero iff $\mathbb{k}\otimes_B M$ and $\mathbb{k}\otimes_A M$ have the same dimension as $\mathbb{k}$ vector spaces, i.e. iff $\text{rank}_A M=\text{rank}_B (M|_B)$.  Further, $\mathbb{k}\otimes_A K$ is zero iff $K$ is a torsion $A$-module.  
