In Atiyah-Macdonald, the following construction is described: Given two rings $A,B$ and $N$ a $B$-module and $f:A\to B$ a homomorphism, we can make $B,N$ into $A$-modules by defining $a\cdot x=f(a)x$. This makes sense to me.
However they go on to describe the $B$-module $N_B= B\otimes_A N$. I can see that, regarded as an $A$-module, this is usually different from $N$, but if we regard it as a $B$-module with the scalar multiplication given by $b'(b\otimes n)=(bb'\otimes n)$ then I get confused.
Exercise 13 asks you to
Show that the homomorphism $g:N\to N_B$ which maps $y$ to $1\otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$
but if $y\to 1\otimes y$ is a $B$-module homomorphism, then we would need to have $by\to b(1\otimes y)=b\otimes y=1\otimes by$, which would make the map an isomorphism. So are we saying that $g$ is an $A$-module homomorphism? But then why do we give $B\otimes_A N$ a $B$-module structure and refer to it as a $B$-module?
