# Annihilator is Submodule of Algebraic Dual

## Theorem

Let $R$ be a commutative ring with unity.

Let $G$ be a module over $R$.

Let $M$ be a submodule of $G$.

Let $G^*$ be the algebraic dual of $G$.

Then the annihilator $M^\circ$ of $M$ is a submodule of $G^*$.

### Corollary

Let $N$ be a submodule of $G^*$.

Let $G^{**}$ be the algebraic dual of $G^*$.

Then the annihilator $N^\circ$ of $N$ is a submodule of $G^{**}$.

## Proof

By definition, $M^\circ$ is a subset of $G^*$.

Recall that by definition of algebraic dual, the elements of $G^*$ are linear transformations from $G$ to the $R$-module $R$.

By Submodule Test, it remains to be shown that:

 $\text {(SM1)}: \quad$ $\ds \forall u, v \in M^\circ: \,$ $\ds u + v$ $\in$ $\ds M^\circ$ $\text {(SM2)}: \quad$ $\ds \forall u \in M^\circ: \forall \lambda \in R: \,$ $\ds \lambda u$ $\in$ $\ds M^\circ$
 $\text {(SM1)}: \quad$ $\ds \forall u, v \in M^\circ: \forall x \in M: \,$ $\ds \map u x + \map v x$ $=$ $\ds 0$ Definition of Annihilator on Algebraic Dual $\ds \leadsto \ \$ $\ds \map {\paren {u + v} } x$ $=$ $\ds 0$ Definition of Pointwise Addition of Linear Transformations $\ds \leadsto \ \$ $\ds u + v$ $\in$ $\ds M^\circ$ Definition of Annihilator on Algebraic Dual

and:

 $\text {(SM2)}: \quad$ $\ds \forall u \in M^\circ: \forall \lambda \in R: \forall x \in M: \,$ $\ds \map {\paren {\lambda u} } x$ $=$ $\ds 0$ Definition of Annihilator on Algebraic Dual $\ds \leadsto \ \$ $\ds \lambda \map u x$ $=$ $\ds 0$ Definition of Linear Transformation $\ds \leadsto \ \$ $\ds \lambda u$ $\in$ $\ds M^\circ$ Definition of Annihilator on Algebraic Dual

$\blacksquare$