Let $R$ be a ring and $M$ be a left $R$ -module. The dual module of $M$ is the right $R$ -module consisting of all module homomorphisms from $M$ into $R$ .

It is denoted by $M^\ast$ . The elements of $M^\ast$ are called 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 $R$ -module $M$ is an $(R,R)$ -bimodule 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 $m \mapsto b(m,-)$ .