Annihilator is Submodule of Algebraic Dual

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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:

\(\text {(SM1)}: \quad\) \(\ds \forall u, v \in M^\circ: \, \) \(\ds u + v\) \(\in\) \(\ds M^\circ\)
\(\text {(SM2)}: \quad\) \(\ds \forall u \in M^\circ: \forall \lambda \in R: \, \) \(\ds \lambda u\) \(\in\) \(\ds M^\circ\)


Indeed:

\(\text {(SM1)}: \quad\) \(\ds \forall u, v \in M^\circ: \forall x \in M: \, \) \(\ds \map u x + \map v x\) \(=\) \(\ds 0\) 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

$\blacksquare$


Sources

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