1
$\begingroup$

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 $\left<A\right>$. Every set $A$ with the property that $\left<A\right>=M$ we call a set of generators of module $M$. If

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

then we denote

$$ \left<a_1, \dots ,a_n\right>=\left<A\right>. $$

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?

$\endgroup$
  • $\begingroup$ 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?) $\endgroup$ – Adam Higgins Mar 3 at 11:49
  • $\begingroup$ Which part don't you understand? $\endgroup$ – Carsten S Mar 3 at 12:18
  • $\begingroup$ 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. $\endgroup$ – dantopa Mar 4 at 1:25
0
$\begingroup$

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 $n\in \mathbb{N}$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Toidi Mar 3 at 12:52
  • $\begingroup$ Yes! For more accuracy you may use $[1],...,[n-1]$ instead of $1,...,n-1$ $\endgroup$ – giannispapav Mar 3 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.