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.)

  • $\begingroup$ Are $A $ and $B $ commutative? I'd assume yes, if this is really from Atiyah-Macdonald, but I'd like to be sure. $\endgroup$ – darij grinberg Aug 30 '19 at 14:06
  • 1
    $\begingroup$ @darijgrinberg In Atiyah-Macdonald, all rings are assumed to be unital and commutative. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.