Linear Transformation of Generated Module

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Theorem

Let $R$ be a ring.

Let $G$ and $H$ be $R$-modules.

Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.

Let $S$ be a generator for $G$.

Suppose that:

$\forall x \in S: \map \phi x = \map \psi x$


Then $\phi = \psi$.


Proof 1

By definitions of generator of module and generated submodule, it follows that:

$\ds G := \bigcap \set { M' \subseteq G : S \subseteq M', \textrm {$M'$ is a submodule of $G$} }$


Set $S' := \set {x \in G: \map \phi x = \map \psi x}$.

From Elements of Module with Equal Images under Linear Transformations form Submodule, it follows that $S'$ is a submodule of $G$.

By hypothesis, $S \subseteq S'$.

It follows that $S'$ is one of the submodules $M'$.

From Intersection is Subset:Family of Sets, it follows that $G \subseteq S'$.

As $S' \subseteq G$, it follows that $S' = G$.

It follows that for all $x \in G$, we have:

$\map \phi x = \map \psi x$

which proves that $\phi = \psi$.

$\blacksquare$


Proof 2

This proof assumes that $R$ is a ring with unity, so $G$ and $H$ become unitary modules.

Let $y \in G$ be arbitrary.

Then by definition of generator, $y$ is the linear combination of elements of $S$:

$\ds y = \sum_{k \mathop = 1}^n \lambda_k a_k$

for $a_1, a_2, \ldots, a_n \in S, \lambda_1, \lambda_2, \ldots, \lambda n \in R$


Then:

\(\ds \map \phi y\) \(=\) \(\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k}\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \lambda_k \map \psi {a_k}\) by hypothesis: $\forall a_k \in S: \map \phi {a_k} = \map \psi {a_k}$
\(\ds \) \(=\) \(\ds \map \psi {\sum_{k \mathop = 1}^n \lambda_k a_k}\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \map \psi y\)

$\blacksquare$


Also see


Sources