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:


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


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


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


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


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

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

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

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

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

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:


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

Then we need to show:


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

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

Proof of Product with Sum of Scalar

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

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

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

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:


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

Then we need to show:


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

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

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.