Basic Results about Modules

Theorem
Let $\left({G, +_G}\right)$ be an abelian group whose identity is $e$.

Let $\left({R, +_R, \times_R}\right)$ be a ring whose zero is $0_R$.

Let $\left({G, +_G, \circ}\right)_R$ be an $R$-module.

Let $x \in G, \lambda \in R, n \in \Z$.

Let $\left \langle {x_m} \right \rangle$ be a sequence of elements of $G$.

Let $\left \langle {\lambda_m} \right \rangle$ be a sequence of elements of $R$ i.e. scalars.

Then:


 * $(1) \quad \lambda \circ e = 0_R \circ x = e$


 * $(2) \quad \lambda \circ \left({- x}\right) = \left({- \lambda}\right) \circ x = - \left({\lambda \circ x}\right)$


 * $\displaystyle (3) \quad \lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$


 * $\displaystyle (4) \quad \left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$


 * $(5) \quad \lambda \circ \left({n \cdot x}\right) = n \cdot \left({\lambda \circ x}\right) = \left({n \cdot \lambda}\right) \circ x$

Proof
From Module: $(1)$, $y \to \lambda \circ y$ is an endomorphism of $\left({G, +_G}\right)$.

From Module: $(2)$, $\mu \to \mu \circ x$ is a homomorphism from $\left({R, +_R}\right)$ to $\left({G, +_G}\right)$.

Proof of Scalar Product with Identity

 * $(1) \quad \lambda \circ e = 0_R \circ x = e$:

This follows from Homomorphism with Cancellable Range Preserves Identity.

Proof of Scalar Product with Inverse

 * $(2) \quad \lambda \circ \left({- x}\right) = \left({- \lambda}\right) \circ x = - \left({\lambda \circ x}\right)$:

This follows from Homomorphism with Identity Preserves Inverses.

Proof of Scalar Product with Sum

 * $\displaystyle (3) \quad \lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$:

This follows by induction from Module: $(1)$, as follows:

For all $m \in \N^*$, let $P \left({m}\right)$ be the proposition:
 * $\displaystyle \lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$

Scalar Product with Sum: Basis for the Induction
$P(1)$ is true, as this just says:
 * $\lambda \circ x_1 = \lambda \circ x_1$

This is our basis for the induction.

Scalar Product with Sum: Induction Hypothesis
Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle \lambda \circ \left({\sum_{k=1}^n x_k}\right) = \sum_{k=1}^n \left({\lambda \circ x_k}\right)$

Then we need to show:


 * $\displaystyle \lambda \circ \left({\sum_{k=1}^{n+1} x_k}\right) = \sum_{k=1}^{n+1} \left({\lambda \circ x_k}\right)$

Scalar Product with Sum: Induction Step
This is our induction step:

So $P \left({n}\right) \implies P \left({n+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m \in \N^*: \lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$

Proof of Product with Sum of Scalar

 * $\displaystyle (4) \quad \left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$:

This follows by induction from Module: $(2)$, as follows.

For all $m \in \N^*$, let $P \left({m}\right)$ be the proposition:
 * $\displaystyle \left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$

Product with Sum of Scalar: Basis for the Induction

 * $P(1)$ is true, as this just says:
 * $\lambda_1 \circ x = \lambda_1 \circ x$

This is our basis for the induction.

Product with Sum of Scalar: Induction Hypothesis

 * Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle \left({\sum_{k=1}^n \lambda_k}\right) \circ x = \sum_{k=1}^n \left({\lambda_k \circ x}\right)$

Then we need to show:


 * $\displaystyle \left({\sum_{k=1}^{n+1} \lambda_k}\right) \circ x = \sum_{k=1}^{n+1} \left({\lambda_k \circ x}\right)$

Product with Sum of Scalar: Induction Step
This is our induction step:

So $P \left({n}\right) \implies P \left({n+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m \in \N^*: \left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$

Proof of Scalar Product with Product

 * $(5) \quad \lambda \circ \left({n \cdot x}\right) = n \cdot \left({\lambda \circ x}\right) = \left({n \cdot \lambda}\right) \circ x$:

First let $n = 0$. The assertion follows directly from result $(1)$ above.

Next, let $n > 0$. The assertion follows directly from results $(3)$ and $(4)$, by letting $m = n$ and making all the $\lambda$'s and $x$'s the same.

Finally, let $n < 0$. The assertion follows from result $(5)$ for positive $n$, result $(2)$, and from Negative Index Law for Monoids.