$R$ : commutative ring with $1$ .

Exercise 369.1(Exercise).

Let $M$ be an $R$ -module. An abelian subgroup $N$ of $M$ is a submodule $⟺$ the inclusion $N ↪ M$ is $R$ -linear.

Exercise 369.2(Exercise).

Let $φ : M \to N$ be a homomorphism. Then $φ$ is an isomorphism $⟺$ $φ$ is bijective.

Aluffi (2009) 5.14, 5.15, 5.16

Every vector space (over a field) has a basis. Not true for arbitrary rings.

Example: Let $R = k [x, y]$ . $I = (x, y)$ is an $R$ -module. $I$ does not have a $R$ -basis:

${x, y}$ clearly generates $I$ , by is not linearly independent: $x y - y x = 0$ .
Any subset of ${x, y}$ does not generate $I$ ; in fact no singleton can generate $I$ .
Repeat $(1)$ for any subset $S \subseteq I$ with $∣ S ∣ ⩾ 2$ .

Definition 369.3(Free module).

An $R$ -module $M$ is said to be free if it has a basis.

Free module generated by. a set. Let $X$ be a set. Then the free module generated by $X$ is the $R$ -module

References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

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