Does $M \otimes_A A \cong M$ and $M \otimes_A I \cong IM$? I am new to the subjects of tensor products and modules. I've been learning about them in my advanced algebra course. 
There are two claims my Professor has made in his notes related to these topics, that I'm having a tough time understanding. Suppose $A$ is a ring, $I$ is an ideal of $A$ and $M$ is an $A$-module. Then, my Professor claimed the following : 


*

*$M \otimes_A A \cong M$

*$M \otimes_A I \cong IM$
How can we define the isomorphism claimed in each of the above ? 
My idea was, for the first isomorphism, to define $\phi: M \otimes_A A \longrightarrow M : m \otimes a \longmapsto ma$. But, if this is the isomorphism, why does $ma$ live in $M$ ? 
Similarly, for the second isomorphism, I thought to define $\phi' : M \otimes_A I \longrightarrow IM : m \otimes i \longmapsto im$. Is this the appropriate isomorphism to get the second claim ? 
Thanks! 
 A: The first claim is correct. $ma$ lives in $M$ if $M$ is a right $A$-module. 
If you only defined left $A$-modules, then $ma$ is an abuse of notation for $am$ (and if $A$ is noncommutative, this matters, so it shoudl be $A\otimes_A M$ - if $A$ is commutative, this is not-so-harmful an abuse)
The second claim is, however, incorect. There is always a map $I\otimes_AM\to M, i\otimes m \mapsto im$ of course, and of course it lands in $IM$ and is surjective, so we have a surjection $I\otimes_AM\to IM$, but in general this map is not injective; and even worse, in some cases there is no injection. In fact, fixing $M$, this map is always injective if and only if $M$ is flat (this is something that you probably don't know yet if you just started modules, so don't worry)
An explicit example is the following : take $A=\mathbb Z$, $M=\mathbb Z/2$ which is naturally a, $A$-module, and $I=2\mathbb Z$. Then $I\cong A$ as $A$-modules, so $I\otimes_A M\cong A\otimes_A M\cong M = \mathbb Z/2$, however, $IM = 2\mathbb Z/2 = 0$, so there is no isomorphism
