dual module DualModule 2006-06-14 19:39:37 2006-10-24 17:05:37 Definition linear functional % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $R$ be a ring and $M$ be a left \PMlinkid{$R$-module}{365}. The {\it dual module} of $M$ is the right \PMlinkid{$R$-module}{365} consisting of all module homomorphisms from $M$ into $R$. It is denoted by $M^\ast$. The elements of $M^\ast$ are called {\it linear functionals}. The action of $R$ on $M^\ast$ is given by $(fr)(m) = (f(m))r$ for $f \in M^\ast$, $m \in M$, and $r \in R$. If $R$ is commutative, then every \PMlinkescapetext{$R$-module} $M$ is an \PMlinkid{$(R,R)$-bimodule}{987} with $rm = mr$ for all $r \in R$ and $m \in M$. Hence, it makes sense to ask whether $M$ and $M^\ast$ are isomorphic. Suppose that $b: M \times M \to R$ is a bilinear form. Then it is easy to check that for a fixed $m \in M$, the function $b(m, -): M \to R$ is a module homomorphism, so is an element of $M^\ast$. Then we have a module homomorphism from $M$ to $M^\ast$ given by $m \mapsto b(m,-)$.