Show that the homomorphism $g: N \rightarrow N_B$ which maps $y$ to $1 \otimes y$ is injective. Let $f:A \rightarrow B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_AN$. Show that the homomorphism $g: N \rightarrow N_B$ which maps $y$ to $1 \otimes y$ is injective.
Comments:
Doing $g(y) = 0$, then $1\otimes y = 0$. I do not know how to show that $y = 0$.
 A: There is a standard adjunction that one can use, by observing mutually inverse operations. We should try and find a left inverse to the map, which will prove the claim.
Let's make this explicit.
A map $\rho:M \to N_A$, where $N_A$ is viewed as an $A$-module via $f$, can be associated to a map $\phi:N_B\to N$ (by extension of scalars) in a very explicit way:
Given $\phi$, we can define $\rho$ by $m \mapsto \rho(m \otimes 1)$.
Likewise, we can also define $\phi$ from $\rho$ by $m \otimes x \mapsto x \cdot\rho(m)$
A candidate for left inverses of your map $N \to N_B$, $y \mapsto y \otimes 1$ might be
$N_B \to N$ given by $y \otimes a \to ay$, but this is the same as $M  \to N_A$ given by $y \mapsto y$ which is identity, so by the above operation, we have a left inverse in the sense that
$$y \mapsto 1 \otimes y \mapsto 1 \cdot y$$
is identity.

Admittedly, the above point of view is not necessary, but I find it illuminating, that extension and restriction by scalars form an adjoint pair. It would suffice to note that our map gives identity and is indeed a module homomorphism, but I think this is how I knew which map to use.

take away: try the map $y \otimes a \mapsto ay$ as a left inverse to your function.
