Here is an exercise of the book commutative algebra by Atiyah and MacDonald (Ex 2.13): Let $ f : A \rightarrow B$ be ring homomorphism and $N$ be a $B$ module. Regarding $N$ as a $A$ module by restricting the scalars, and form the $B$ module $N_{B} = B \otimes N$ (tensor over $A$). Then the homomorphism $g : N \rightarrow N_{B}$ which maps $y$ to $ 1 \otimes y $ is injective and $g(N)$ is a direct summand of $N_{B}$. i.e. $N_{B} = L \oplus g(N)$ for some $B$ module $L$.

I think I am not understanding this statement in the following way: let's take $B$ to be any ring containing a field $k$, let's $N$ be any $B$ module, by restricting the scalars to $k$, it becomes $k$ module hence free module therefore projective, then $N_{B}$ becomes projective $B$-module therefore any direct summand is projective which implies that $N$ is projective $B$ module, which is completely absurd because $N$ was any $B$ module. Where is the flaw? Any help would be great.

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    $\begingroup$ It seems the given map $g$ is not $B$-linear (it is only $A$-linear) since the $B$-module structure on $N_{B}$ is by $b' \cdot (b \otimes y) = b'b \otimes y$, which may not be equal to $b \otimes b'y$. (The map $p$ in the hint to the exercise is $B$-linear though.) Maybe they meant "direct summand as $A$-modules"? $\endgroup$ – Minseon Shin Jun 12 '19 at 8:51
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    $\begingroup$ That means that $g(N)$ is direct summand as $A$ module not as a $B$ module.Am I correct? $\endgroup$ – Sunny Jun 12 '19 at 11:49
  • $\begingroup$ Yes, that's what I mean. $\endgroup$ – Minseon Shin Jun 12 '19 at 21:35
  • $\begingroup$ See also math.stackexchange.com/questions/2025812/… $\endgroup$ – Minseon Shin Aug 30 '19 at 19:44