Simple, Cyclic and Projective Modules

Question

I have been taking a course in Homological Algebra and revisiting the lecture notes for some reason. There was a non-answered question about showing two non zero modules A,B over a non trivial unital ring(in fact upper triangular matrices) are projective modules. I won't explain what these modules are because it requires a lot of work.

I thought that if there is a relation between being cyclic and/or simple AND projective then I can use that. However, I couldn't find any such relation.

Is there any relation between cyclic and/or simple modules and projective modules?

Answer 1

All modules are nonzero in the following discussion.

• Simple modules are cyclic. In fact, every nonzero element of a simple module can act as a generator. Indeed, let $M$ be a simple module and $x \in M$ be a nonzero element of $M$. Let's consider the submodule $Rx$ generated by $x$ in $M$. Since $M$ is simple and $Rx \neq 0$, it has to be $M$, i.e. $M=Rx$. This shows that $M$ is cyclic.

• Every cyclic module can be written as a quotient of the regular module $R$ (i.e. $R$ is viewed as an $R$-module.) In fact, let $M=Rx$ for some $x \in M$, then consider the annilihaltor of $x$, i.e. $$ \mathrm{ann}_{R}(x) = \{ r \in R : rx=0\}. $$ Then it is the kernel of the left multiplication map $$ \ell_{x}: R \rightarrow M=Rx;\quad r \mapsto rx. $$ This is clearly surjective. Hence by the isomorphim theorem, we have $$ M \cong R/\mathrm{ann}_{R}(x). $$ Hence if $x$ is a torsion free element (i.e. $\mathrm{ann}_{R}(x)=0$), then $M$ is a free $R$-module of rank 1.

• Free modules are projective. (This is shown in most of the homological algebra textbooks.)

This is what I can think of among simple modules, cyclic module and projective modules. Sorry for any possible mistakes and misleadings. :)

EDIT: Glad to see that this answer is helpful. As a reply to the comment, I shall add a few things.

Item 1: There are examples of cyclic modules that are not simple. Examples are rich when we go back to (finitely generated) $\mathbb{Z}$-modules, i.e. abelian groups. This case is much more clearer since we can apply both the group-theoretic facts and the structure theory of finitely generated modules over PID. For finite $\mathbb{Z}$-modules (i.e. finite abelian groups, or say finitely generated abelian groups with no free parts), it is inspiring to consider the following problems:

• Classify all cyclic finite $\mathbb{Z}$-modules and all its possible $\mathbb{Z}$-submodules;
• Classify all simple (or irreducible) finite $\mathbb{Z}$-modules;
• Classify all indecomposable finite $\mathbb{Z}$-modules;
• Classify all semisimple finite $\mathbb{Z}$-modules.

If one hopes to get more familiar to examples, one can substitude the $\mathbb{Z}$-modules in above questions to $F[X]$-modules to invoke linear algebra stuff. (Here $F$ is a field, maybe one can assume $F$ is algebraically closed of characteristic 0, or say $F =\mathbb{C}$. I haven't consider these questions over $F[X]$, but it seems quite illustrating.)

Still feel not quite enough after tried the above two cases? Then what about considering a family of linear operators acting on the vector spaces? Say we have two commuting linear operators on the vector space $V$, then we can shift to consider $F[X,Y]$-module. This is quite complicated (and I haven't tried by hand and I'm not suggesting you to do this and the follows in this paragraph). If we have a group of linear operators, say a finite group $G$ acting on $V$, then we are actually considering the $F[G]$-modules. With a little bit modification, we now step into the field of group representation theory. There, classifying irreducible $F[G]$-modules and see how a reducible one decomposes is the main topic in a first course on representation theory of finite groups. For infinite groups, things can even be much more complicated.

You mentioned modules over upper triangular matrices. Then it maybe relevant to the case that a family of upper-triangulariable operators acting on a vector space? (Actually in finite dim'l vector spaces over $\mathbb{C}$, all linear operators are upper-triangulariable if I haven't got it wrong.)

Sorry I have said so many irrelevant stuff to the question. Yet it excited me so much when I first realized such things before.

Or in general, one may prove that

• A projective module over a PID is precisely a free module;
• An injective module over a PID is precisely a divisible module (requires Baer's criterion);
• A flat module over a PID is precisely a torsion-free module.

Item 2: Actually there is a description of projective modules: An $R$-module $M$ is projective iff it is a direct summand of a free module $R^{\oplus I}$. This may help in some cases and might be usefull when describing the relation among simple/cyclic/projective modules.

Answer 2

I thought that if there is a relation between being cyclic and/or simple AND projective then I can use that.

1. A simple module is always cyclic ( every nonzero element generates it.)

2. A projective cyclic module need not be simple (for example $R_R$ for any ring that isn’t a division ring.)

3. A simple module need not be projective. (For example, $\mathbb Z/2\mathbb Z$ is not projective as a $\mathbb Z$ module.)

I think this takes care of all the possible mixtures of relationships that you could test. There is very little to say positively.