# Submodule generated by a set

I don't understand this definition. Can someone explain it to me and give some simple examples?

Let $$R$$ be a ring, $$M$$ a left $$R$$-module and $$A \subset M$$ some set. The least (in terms of inclusion) submodule of module $$M$$ containing $$A$$ (i.e the intersection of all submodules of module $$M$$ containing $$A$$) we call the submodule generated by $$A$$ and mark $$\left$$. Every set $$A$$ with the property that $$\left=M$$ we call a set of generators of module $$M$$. If

$$A=\{a_1, \dots ,a_n\},$$

then we denote

$$\left=\left.$$

We say that a module is finitely generated (cyclic) if there exists a finite (with one element) set of generators.

Are there any equivalent definitions?

• You have a few misconceptions here. A module is said to be cyclic if is has a generating set consisting of a single element. Think of cyclic groups as examples of cyclic $\mathbb{Z}$-modules. A module is said to be finitely generated if there exists a finite generating set of any size. So cyclic is a much stronger condition than finitely generated. An example of a non-finitely generated module is the Abelian group $\mathbb{Q}$ as this does not have a generating set of any finite size (can you prove this?) – Adam Higgins Mar 3 at 11:49
• Which part don't you understand? – Carsten S Mar 3 at 12:18
• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. – dantopa Mar 4 at 1:25

## 1 Answer

Example:

Let $$R=\mathbb{Z}$$ and $$M$$ an abelian group, then $$M$$ is a $$\mathbb{Z}-$$module. So $$M$$ is finitely generated $$\mathbb{Z}-$$module (or abelian group) iff there is a $$A\subset M$$ with $$|A|<\infty$$ s.t $$M=\langle A\rangle=\langle x_1,...,x_n\rangle$$ which means that every $$m\in M$$ is $$m=r_1x_1+...+r_nx_n, \quad r_i\in \mathbb{Z}$$

Equivalent definition:

$$M$$ is finitely generated $$R-$$module if and only if there is a surjective $$R$$-linear map: $$f:R^n\to M$$ for some $$n\in \mathbb{N}$$

• So even more simplifying, lets consider $M=Z_n$. We can write $\left<A\right>=\left<1,...,n-1\right>$ or even $\left<A\right>=\left<1\right>$ and both statements are correct? – Toidi Mar 3 at 12:52
• Yes! For more accuracy you may use $,...,[n-1]$ instead of $1,...,n-1$ – giannispapav Mar 3 at 12:54