# Module over restriction of scalars [duplicate]

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.

• 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"? – Minseon Shin Jun 12 '19 at 8:51
• That means that $g(N)$ is direct summand as $A$ module not as a $B$ module.Am I correct? – Sunny Jun 12 '19 at 11:49
• Yes, that's what I mean. – Minseon Shin Jun 12 '19 at 21:35
• – Minseon Shin Aug 30 '19 at 19:44