# Generated Submodule may not equal Set of Linear Combinations

## Theorem

Let $\struct { R, +_R, \times_R }$ be a ring.

Let $\struct { M, +_M, \circ }_R$ be an $R$-module.

Let $S$ be a subset of $M$.

Let $H_1$ be the submodule generated by $S$.

Let $H_2$ be the set of all linear combinations of elements of $S$.

Then it is possible to select $\struct { R, +_R, \times_R }$, $\struct { M, +_M, \circ }_R$, and $S$ such that:

$H_1 \ne H_2$

## Proof

Let $R = \set {2 k : k \in \Z}$ be the set of all even integers.

From Integers form Commutative Ring, it follows that $\Z$ is a ring.

From Ideal of Ring/Examples/Set of Even Integers, it follows that $R$ is an ideal of $\Z$.

From Ideal is Subring, it follows that $\struct {R, +, \times}$ is a ring, where $+$ denotes integer addition, and $\times$ denotes integer multiplication.

Set $M := R$.

It follows that $\struct {R, +, \times }_R$ is a module over $R$, where the scalar multiplication $\times$ is defined as the integer multiplication $\times$ in $R$.

That is, $R$ is considered as a module over itself.

Set $S := \set{ 2 }$.

Let $H_1$ be the submodule generated by $S$, which means that:

$\ds H_1 := \bigcap \set { M' \subseteq R : \set {2} \subseteq M', \textrm {$M'$is a submodule of$R$} }$

Let $M'$ be a submodule of $R$ such that $2 \in M'$.

We show that $M' = R$.

From Congruence Modulo Integer is Equivalence Relation, it follows that all $x \in \Z$ is of the form $x := 4 k' + r$ for some $k \in \Z, r \in \set{ 0,1,2,3 }$.

Suppose $x \in R$ such that $x = 4 k'$ for some $k' \in \Z$

Then:

$4 k' = 2 k' \times 2$

As $M'$ is closed for scalar product, it follows that $4 k' \in M'$.

Suppose $x \in R$ such that $x = 4 k' + 2$ for some $k' \in \Z$.

As $M'$ is closed for addition, and $2 \in M'$, it follows that $4 k' + 2 \in M'$.

If instead $x = 4 k' + 1$ or $x = 4 k' + 3$, then $x$ cannot be an even integer, so $x \notin R$.

It follows that $R \subseteq M'$.

As $M'$ is a submodule of $R$, it follows that $M' \subseteq R$.

As we have $M' = R$, it follows that:

$\bigcap \set { M' \subseteq R : \set {2} \subseteq M', \textrm {$M'$is a submodule of$R$} } = \bigcap \set {R} = R$

Let $H_2$ be the set of all linear combinations of elements of $\set {2}$, which means that:

$H_2 = \ds \set {\sum_{i \mathop = 1}^n \lambda_i \times 2 : n \in \N_{\ge 1}, \lambda_1, \ldots, \lambda_n \in R}$

With $\ds \sum_{i \mathop= 1}^n \lambda_i \times 2 \in H_2$, it follows that:

 $\ds \sum_{i \mathop= 1}^n \lambda_i \times 2$ $=$ $\ds \sum_{i \mathop= 1}^n \paren{ 2 k_i } \times 2$ for some $k_1, \ldots, k_n \in \Z$ $\ds$ $=$ $\ds 4 \sum_{i \mathop= 1}^n k_i$

As $\ds \sum_{i \mathop= 1}^n k_i \in \Z$, it follows that:

$H_2 = \set { 4 k : k \in Z}$

Hence, $H_1 \ne H_2$.

$\blacksquare$