# Exercise 2.13 in Atiyah-Macdonald

Let $f \colon A \to B$ 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 $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.
[Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]

My idea was to define an isomorphism \begin{align*} \phi \colon N_B & \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y & \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}

It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$b' . (b \otimes y) = (b'b) \otimes y$$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y)$$ (where $\neq$ means "not equal in general").

However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.

Should I take the $B$-module structure of $N_B$ to be something else, for example $$b'.(b \otimes y) = b \otimes (b'.y)$$ or should I show the result only for $A$-modules?

I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)

• Are $A$ and $B$ commutative? I'd assume yes, if this is really from Atiyah-Macdonald, but I'd like to be sure. – darij grinberg Aug 30 '19 at 14:06
• @darijgrinberg In Atiyah-Macdonald, all rings are assumed to be unital and commutative. – sera Aug 30 '19 at 14:11

Recall that $$g(N)$$ is a direct summand of $$N_B$$, if it is a submodule of $$N_B$$ and there exists a direct complement $$M \leq N_B$$ such that $$g(N) \oplus M = N_B$$. Thus it should be mentioned whether $$g(N)$$ is a direct summand of $$N_B$$ as $$A$$-modules or $$B$$-modules.
As you pointed out, $$g \colon N \ni y \mapsto 1 \otimes y \in N_B$$ is not a homomorphism of $$B$$-modules in general. For example, let $$A=\mathbb{R}$$, $$B=N=\mathbb{C}$$ and $$f$$ be the inclusion. Then we get $$i \otimes 1 \neq 1 \otimes i$$ in $$\mathbb{C} \otimes _ \mathbb{R} \mathbb{C}$$. Moreover, here $$g(N)$$ is not a direct summand of $$N_B$$ as $$B$$-modules, since $$g(N)=g(\mathbb{C} \otimes_\mathbb{R} \mathbb{C})$$ is not a $$B$$-submodule of $$N_B$$. To see this, note that $$\mathbb{C} \otimes _\mathbb{R} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$$ as $$\mathbb{C}$$-modules via $$x \otimes y \mapsto (xy, x \bar{y})$$. Under this identification, $$g(N)$$ is isomorphic (as $$\mathbb{R}$$-modules) to $$\{(y, \bar{y}):y \in \mathbb{C} \} \leq \mathbb{C} \times \mathbb{C}$$, which is not a $$\mathbb{C}$$-submodule.
Here is my approach: It is easy to show $$g$$ and $$p$$ are $$A$$-module homomorphisms satisfying $$p \circ g = id_N$$. In particular, $$g$$ is injective. Now consider the short exact sequence of $$A$$-modules: $$0 \longrightarrow N \overset{g}\longrightarrow N_B \longrightarrow N_B/g(N) \longrightarrow 0$$ Since $$p$$ is a left inverse of $$g$$, this exact sequence splits. Hence $$N_B = \ker(p) \oplus \text{im} (g)$$ as $$A$$-modules.