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$F (A)$ is a $(B, A)$ -bimodule for every additive functor $F$

Statement

Let $A$ and $B$ be unital rings and let $F : A - M o d \to B - M o d$ be an additive functor (Additive functor). On the left $B$ -module $M = F (A)$ define, for $a \in A$ and $m \in M$ ,

$m a : = F (r_{a}) (m),$

where $r_{a} : A \to A$ , $r_{a} (x) = x a$ , is right multiplication, a left $A$ -linear endomorphism of $A$ . Together with the given left $B$ -action this makes $M$ a $(B, A)$ -bimodule ( $(S, R)$ -bimodules and commuting left and right scalar actions): the right $A$ -action satisfies the unit, associativity and distributivity laws of Unital left and right modules over a ring; unqualified module means left module and commutes with the left $B$ -action. No commutativity of the rings is assumed and no choice is used.

Facts & Assumptions

Given: Unital rings $A$ and $B$ , an additive functor $F : A - M o d \to B - M o d$ , and the left $B$ -module $M = F (A)$ .

[F1]

A right $R$ -module is an abelian group with an action $(m, r) \mapsto m r$ satisfying the right-handed analogues of the left-module axioms, in particular $m 1 = m$ , $(m + m^{'}) a = m a + m^{'} a$ , $m (a + a^{'}) = m a + m a^{'}$ and $(m a) a^{'} = m (a a^{'})$ (Unital left and right modules over a ring; unqualified module means left module). Multiplication in a ring satisfies $(c x) a = c (x a)$ and $x (a + a^{'}) = x a + x a^{'}$ .

[F2]

An additive functor satisfies $F (f + g) = F f + F g$ for every parallel pair of morphisms $f, g$ (Additive functor).

[F3]

A functor satisfies $F (1_{X}) = 1_{F X}$ and $F (g \circ f) = F g \circ F f$ (Covariant functor, identity functor, composite functor, and contravariant functor).

[F4]

An $(S, R)$ -bimodule is an abelian group that is a left $S$ -module and a right $R$ -module whose two actions commute ( $(S, R)$ -bimodules and commuting left and right scalar actions).

Proof

technique · direct

1.1F1F3

For $a \in A$ the map $r_{a} : A \to A$ , $r_{a} (x) = x a$ , is a left $A$ -module endomorphism, because $r_{a} (x + y) = r_{a} (x) + r_{a} (y)$ and $r_{a} (c x) = (c x) a = c (x a) = c r_{a} (x)$ by the ring laws. Hence $F (r_{a}) : M \to M$ is a $B$ -linear, in particular additive, endomorphism, and $m a : = F (r_{a}) (m)$ defines a map $M \times A \to M$ with $(m + m^{'}) a = m a + m^{'} a$ for all $m, m^{'} \in M$ .

1.2F1F3

Unit law: $r_{1} = {i d}_{A}$ because $x 1 = x$ , so $m 1 = F ({i d}_{A}) (m) = F (1_{A}) (m) = 1_{M} (m) = m$ for every $m \in M$ .

1.3F1F3

Associativity: $r_{a a^{'}} = r_{a^{'}} \circ r_{a}$ because $x (a a^{'}) = (x a) a^{'}$ , so functoriality gives $F (r_{a a^{'}}) = F (r_{a^{'}}) F (r_{a})$ and hence $m (a a^{'}) = (m a) a^{'}$ for all $m \in M$ and $a, a^{'} \in A$ .

1.4F1F2

Additivity in the ring variable: $r_{a + a^{'}} = r_{a} + r_{a^{'}}$ because $x (a + a^{'}) = x a + x a^{'}$ , and additivity of $F$ gives $F (r_{a + a^{'}}) = F (r_{a}) + F (r_{a^{'}})$ , so $m (a + a^{'}) = m a + m a^{'}$ .

2.1F4givenstep 1.1

Commutation with the left $B$ -action: for $b \in B$ and $m \in M$ we have $b (m a) = b (F (r_{a}) (m)) = F (r_{a}) (b m) = (b m) a$ , since the morphism $F (r_{a})$ of $B - M o d$ is $B$ -linear.

3.1F1F4step 1.1step 1.2step 1.3step 1.4step 2.1∎

Steps 1.1-1.4 make $M$ a right $A$ -module for the assignment $(m, a) \mapsto m a$ , and step 2.1 shows that this right $A$ -action commutes with the given left $B$ -action; by [F4] the abelian group $M$ is a $(B, A)$ -bimodule.

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Kamensky, Non-Commutative Algebra (BGU course notes, Spring 2017), §5.1, Theorem 5.1.43, Proposition 5.1.40, Lemma 5.1.46, Corollaries 5.1.48-5.1.49 (standard reference, not scraped)
A. Nyman and S. P. Smith, A Generalization of Watts's Theorem: Right Exact Functors on Module Categories, arXiv:0806.0832, Theorem 1.1-1.2, Propositions 3.2-3.3, Lemma 3.4 (standard reference, not scraped)

F(A)F(A) is a (B,A)(B,A)-bimodule for every additive functor FF