# Free modules over commutative ring (possibly without unity) where free means having a LI spanning set

Let us define free module over a ring (possibly without unity) as:

Def: M is said to be free module over ring R (possibly without unity) if there exist X subset of M such that X is LI and spans M. Any such X is called basis of M.

Def: If M free module over a ring R. Cardinality of any basis is called rank of M (we are allowing M to have two different ranks)

Def: R is said to have IBN property if any free module over R has fixed rank.

Question: Does following rings have IBN property

1. Commutative ring
2. Commutative ring R with property that there exist r in R such that rs never 0 for all s≠0
3. ID without unity

In other words can we find free module over R with two different ranks where R one of above rings?

Motivation for question: Any commutative ring with unity or ID has IBN property.

Remark: This definition of free module is not at all same to what is called free object in the category of modules

Are there notes of some professor who have taught this definition of free module?

• Dunno about rings without unity. However: 1) commutative rings with unity always have IBN (see Rotman's books), 2) does not make sense to me, perhaps you mean $s\neq0$, or something else is missing, i.e. you want a division ring? Division rings (with unity) always have IBN as well. – M.G. Jan 2 '16 at 9:57
• Yes commutative rings with unity always have IBN. This I know. Sorry for mistake, yes I mean s≠0. But how does it imply divison ring. And question is for rings without unity. @July – Sushil Jan 2 '16 at 12:04
• It does not imply division ring, I just wasn't sure what the missing part in that statement was. – M.G. Jan 2 '16 at 12:45
• I'm having trouble with your non-stated definition of linear independent set $X$ that spams $M$. As a test example, take $R$ to be the subset $\{0,2\}$ of $\mathbb{Z}/4$ (it is a ring without unity), $M=\mathbb{Z}/4$ and $X=\{1\}$. Would you say that $X$ is a free generating set of $M$? If not, why not? It does generate $M$ as an abelian group... I think it would help to state both the definition of LI and of spamming set. – Pierre-Guy Plamondon Jan 5 '16 at 11:33
• @Pierre-GuyPlamondon sorry just recognised what you said. Yes Z/4 is free mofule over {0,2} with basis {1}. But still definitions are well defined and questions are still well defined. – Sushil Jan 6 '16 at 12:32

## 1 Answer

If I understand correctly, you say that an $R$-module $M$ is free if there exists a subset $X$ of $M$ such that the map $$R^{(X)}\to M: (r_x)_{x\in X}\mapsto \sum_{x\in X}r_x\cdot x$$ is bijective.

In that case, the following pretty much answers all your questions: If $R$ is commutative and if there exists a free $R$-module, then $R$ has a unity.

Indeed, assume that $M$ is a free $R$-module with basis $X$. Fix $x_0\in X$; there exists a family of elements of $R$ with finite support $(r_x)_{x\in X}$ such that $$x_0 = \sum_{x\in X}r_x\cdot x.$$

For any non-zero $s\in R$, $sx_0$ is non-zero, so $$- sx_0 + \sum_{x\in X}sr_x\cdot x = 0.$$

By the injectivity of the map above, we get $sr_x=0$ whenever $x\neq x_0$, and $sr_{x_0}-s=0$. Thus, for all $s\in R$, we have that $sr_{x_0} = s$; in other words, $r_{x_0}$ is a unity in $R$.

• But you using R is ID or there exist x0 in R such that s.x0 never 0 for all s≠0. Am I right?(or does it follow from definition of free module only) – Sushil Jan 13 '16 at 5:25
• $x_0$ is in $M$, not in $R$; $sx_0\neq 0$ follows from your definition of free module. This is one of the reasons I was asking for a precise definition. – Pierre-Guy Plamondon Jan 13 '16 at 7:48
• I had meant same as you mentioned in beginning of your question. But fortunately it implies R has unity in case of commutative rings – Sushil Jan 13 '16 at 11:30