Preimage 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 $N$ be a submodule of $H$.

Then $\phi^{-1} \sqbrk N$ is a submodule of $G$.


Proof

Let $M = \phi^{-1} \sqbrk N$ be the preimage of $N$ under $\phi$.

Aiming for a contradiction, suppose $M$ is not a submodule of $G$.

This means that $M$ does not fulfil all the module axioms.


First suppose that:

$(1): \quad \struct {M, +_G}$ is not a subgroup of $G$.

Then $M$ is not a group.

Then by the One-Step Subgroup Test:

$\exists x, y \in M: x +_G \paren {-y} \notin M$

Then:

\(\ds \exists x, y \in M: \, \) \(\ds x +_G \paren {-y}\) \(\notin\) \(\ds M\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {x +_G \paren {-y} }\) \(\notin\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x +_H \map \phi {-y}\) \(\notin\) \(\ds N\)

But by definition of $\phi$, both $\map \phi x \in N$ and $\map \phi {-y} \in N$.

Hence $\struct {N, +_G}$ is not closed.

By Group Axiom $\text G 0$: Closure as applied to $N$, it follows that $N$ is not a group.

Hence $N$ is not a subgroup of $H$.

From this contradiction it follows that $(1)$ is false.

Thus $\struct {M, +_G}$ is a subgroup of $G$.

$\Box$


We have by hypothesis that $M$ is not a submodule of $G$.

We have shown that $\struct {M, +_G}$ is a subgroup of $G$.

Hence it must be the case that $M$ is not closed for scalar product:

\(\text {(2)}: \quad\) \(\ds \exists \lambda \in R, x \in M: \, \) \(\ds \lambda \circ_G x\) \(\notin\) \(\ds M\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {\lambda \circ_G x}\) \(\notin\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds \lambda \circ_H {\map \phi x}\) \(\notin\) \(\ds N\)

But by definition of $\phi$, both $\map \phi x \in N$.

Hence $N$ is not closed for scalar product.

From this contradiction it follows that $(2)$ is false.

Thus $\struct {M, +_G}$ is closed for scalar product.

$\Box$


We have shown that $M$ is a subgroup of $G$ which is closed for scalar product.

That is, $M$ is a submodule of $G$.

$\blacksquare$


Sources

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