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 elements of $R$ i.e. scalars.

Then:


 * 1) $$\lambda \circ e = 0_R \circ x = e$$;
 * 2) $$\lambda \circ \left({- x}\right) = \left({- \lambda}\right) \circ x = - \left({\lambda \circ x}\right)$$;
 * 3) $$\lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$$;
 * 4) $$\left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$$;
 * 5) $$\lambda \circ \left({n \cdot x}\right) = n \cdot \left({\lambda \circ x}\right) = \left({n \cdot \lambda}\right) \circ x$$.

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

From Module: VS 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
$$\lambda \circ e = 0_R \circ x = e$$:

This follows from Homomorphism with Cancellable Range Preserves Identity.

Proof of Scalar Product with Inverse
$$\lambda \circ \left({- x}\right) = \left({- \lambda}\right) \circ x = - \left({\lambda \circ x}\right)$$:

This follows from Homomorphism with Identity Preserves Inverses.

Proof of Scalar Product with Sum
$$\lambda \circ \left({\sum_{k=1}^m x_k}\right) = \sum_{k=1}^m \left({\lambda \circ x_k}\right)$$:

This follows by induction from Module: VS 1.

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

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.

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:

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

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({n}\right) \Longrightarrow P \left({n+1}\right)$$ 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
$$\left({\sum_{k=1}^m \lambda_k}\right) \circ x = \sum_{k=1}^m \left({\lambda_k \circ x}\right)$$:

This follows by induction from Module: VS 2.

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

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.

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:

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

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({n}\right) \Longrightarrow P \left({n+1}\right)$$ 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
$$\lambda \circ \left({n \cdot x}\right) = n \cdot \left({\lambda \circ x}\right) = \left({n \cdot \lambda}\right) \circ x$$:

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.