# 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”.

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. 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. 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 $\{e_{1},\ldots,e_{n}\}$ be a $k$-basis of $M$. Of course $\{e_{1}^{*},\ldots,e_{n}^{*}\}$ 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:

 $\xymatrix{X\ar[r]^{f}\ar[d]_{\tau_{X}}&Y\ar[d]^{\tau_{Y}}\\ DD(X)\ar[r]^{DD(f)}&DD(Y)}$

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. $\square$

Title standard duality on modules over algebras StandardDualityOnModulesOverAlgebras 2013-12-11 15:25:39 2013-12-11 15:25:39 joking (16130) joking (16130) 5 joking (16130) Theorem msc 16S99 msc 20C99 msc 13B99