standard duality on modules over algebras

Let $k$ be a field and let $A$ be an associative unital algebra. Throughout we will assume that all $A$-modules over $k$ are unital. If $M$ is a right $A$-module, then the space of all linear mappings

$${\mathrm{Hom}}_k(M,k)$$

can be equipped with a left $A$-module structure as follows: for any $f\in {\mathrm{Hom}}_k(M,k)$ and $a\in A$ put

$$(af)(x)=f(xa).$$

Note that action direction need to be reversed, because

$$(abf)(x)=(bf)(xa)=f(xab).$$

Analogously ${\mathrm{Hom}}_k(-,k)$ takes left $A$-modules to right $A$-modules. Also this action is compatible with functoriality of ${\mathrm{Hom}}_k(-,k)$, which means that it takes $A$-homomorphisms to $A$-homomorphisms. In particular we obtain a (contravariant) functor from category of left (right) $A$-modules to category of right (left) $A$-modules. Obviously $\mathrm{Hom}$ does not change the dimension of spaces, so we have well defined functors

$$D:\mathrm{mod}A\to A\mathrm{mod}$$

$$D:A\mathrm{mod}\to \mathrm{mod}A$$

which are restrictions of $\mathrm{Hom}$ (here $\mathrm{mod}$ means finite dimensional modules left/right modules) and are known in literature as ,,standard dualities”.

Proposition. Both $D$’s are quasi inverse dualities of categories.

Proof. Let $M$ be a finite dimensional $A$-module. We need to define a natural isomorphism between $M$ and $DD(M)$. Indeed, define

$${\tau }_M:M\to DD(M);$$

$${\tau }_M(m)(\alpha )=\alpha (m).$$

We will show that each $\tau$ is an isomorphism.

1. First we will show that $\tau$ is a monomorphism. Assume that ${\tau }_M(m)=0$ for nonzero $m\in M$. This is if and only if $\alpha (m)=0$ for every linear mapping $\alpha :M\to k$. But $m$ is nonzero, so there is a basis of $M$ (as linear space) which contains $m$. In particular there is a linear mapping $f:M\to k$ such that $f(m)=1$. Contradiction. Thus $m=0$, which completes this part.

2. $\tau$ is an epimorphism. Indeed, let $F:D(M)\to k$ be a linear mapping. We need to show, that there is $m\in M$ such that

   $$F(\alpha )=\alpha (m)$$

   for any $\alpha \in D(M)$. Since $M$ is finite dimensional, then let $\{e_1,\mathrm{\dots },e_n\}$ be a $k$-basis of $M$. Of course $\{e_1^{*},\mathrm{\dots },e_n^{*}\}$ is a basis of $D(M)$, where $e_i^{*}$ is given by $e_i^{*}(e_j)=1$ if $i=j$ and $e_i^{*}(e_j)=0$ otherwise. Define

   $${\lambda }_i=F(e_i^{*})$$

   and put

   $$m=\sum _{i=1}^n{\lambda }_i\cdot e_i.$$

   We leave it as a simple exercise, that $\tau (m)=F$.

What remains is to prove, that $\tau$ is natural. Consider an $A$-homomorphism $f:X\to Y$. We need to show that the following diagram commutes:

Indeed, if $x\in X$, then let $F={\tau }_X(x)$. We have that

$$DD(f)(F)=F\circ D(f)$$

and evaluating this at $\alpha \in D(M)$ we have

$$F(D(f)(\alpha ))=F(\alpha \circ f)=(\alpha \circ f)(x)=\alpha (f(x))={\tau }_Y(f(x))(\alpha ).$$

In particular we obtain that

$$DD(f)({\tau }_X(x))={\tau }_Y(f(x))$$

which means that

$$DD(f)\circ {\tau }_X={\tau }_Y\circ f$$

which completes the proof. $\mathrm{\square }$

Title	standard duality on modules over algebras
Canonical name	StandardDualityOnModulesOverAlgebras
Date of creation	2013-12-11 15:25:39
Last modified on	2013-12-11 15:25:39
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Theorem
Classification	msc 16S99
Classification	msc 20C99
Classification	msc 13B99