Image of Submodule under Linear Transformation is Submodule

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Theorem

Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.

Let $\phi: G \to H$ be a linear transformation.


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

Then $\phi \sqbrk M$ is a submodule of $H$.


Proof

Let $N = \phi \sqbrk M$ be the image set of $M$ under $\phi$.


By definition, a linear transformation $\phi: G \to H$ is, in particular, a (group) homomorphism from the group $\struct {G, +_G}$ to the group $\struct {H, +_H}$.

We have by hypothesis that $M$ is a submodule of $G$.

So from Elements of Submodule form Subgroup, $M$ forms a subgroup of $G$.

From Group Homomorphism Preserves Subgroups, $N$ is therefore a subgroup of $H$.

$\Box$


It remains to be shown that $N$ is closed for scalar product:

$\forall \lambda \in R, x \in N: \lambda \circ_H x \in N$


As $M$ is a submodule of $G$, $M$ itself is closed for scalar product:

$\forall \lambda \in R, y \in N: \lambda \circ_G y \in M$


We have:

\(\ds \forall n \in N: \exists y \in M: \, \) \(\ds x\) \(=\) \(\ds \map \phi y\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds \forall \lambda \in R: \, \) \(\ds \lambda \circ_H x\) \(=\) \(\ds \lambda \circ_H \map \phi y\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ_G y}\) Definition of Morphism Property applied to $\phi$
\(\ds \leadsto \ \ \) \(\ds \lambda \circ_H x\) \(\in\) \(\ds \phi \sqbrk M\) as $\lambda \circ_G y \in M$
\(\ds \leadsto \ \ \) \(\ds \lambda \circ_H x\) \(\in\) \(\ds N\) Definition of $N$

$\blacksquare$


Sources

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