# For which modules over commutative rings has the set of all its bases exactly one element?

Question. Is there a reasonable characterization of which modules over commutative rings have exactly one basis?

Remarks.

• This is probably a rather trivial curiosity, yet it comes up naturally in something else I am doing, and, while it is not central to that, it irritates me that I don't see an answer. I have decided to leave it by the wayside, and hope this is of some instructional value to someone else.

• Trivial examples of modules with set-wise precisely one basis are: $\{0\}$ for any commutative ring (the unique basis is $\emptyset$), and also $\mathbb{Z}/2\mathbb{Z}$ as module over itself (the unique basis is $\{ 1 + (2)_{\mathbb{Z}}\}$).

• The obvious thing to try to answer this is of course to argue thus: modules are abelian groups with a specified ring-action, and a group is not empty, $M$ has at least one element. If $M$ has exactly one element, then $M=\{0\}$, with $0\in R$, and this $M$ has the unique basis $\{\}$. If on the contrary $M$ has more than one element, then let $B$ be any $R$-basis of $M$. Since $M$ is not $\{0\}$, $B$ is not empty. For any $S\subseteq R$ let $(a)_{S} :=\{r\cdot a\colon r\in S\}$ denote the principal ideal generated by $a$. If $R=\mathbb{Z}/2$ and $M = \{ 0+(2)_{\mathbb{Z}} , 1+(2)_{\mathbb{Z}} \}$, then $\{ 1+(2)_{\mathbb{Z}} \}$ is the unique $R$-basis of M$. If$M$is a free$R$-module of$R$-rank$\geq 2$, consider any$R$-basis$B = \{ b_i\colon i\in\mathrm{rank}(M)\} $of$M$. Then by definition$M=\bigoplus_{i\in\mathrm{rank}(M)} \langle b_i\rangle_R$and now note that$\{b_0+b_1 , b_1\}\cup B\setminus\{b_0,b_1\}$is distinct as a set from$B$(since$b_0+b_1=b_0$implies$b_1=0$which is impossible), and is a basis of$M$again (since$b_0 = (b_0+b_1)-b_1$shows it is again a generating system, and since$\lambda_0\cdot (b_0 + b_1) + \lambda_1\cdot b_1 + \sum_{i\in\mathrm{rank}(M)\setminus2} b_i= 0$, with cofinitely many summands zero, implies$\lambda_0\cdot b_0 + (\lambda_0+\lambda_1)\cdot b_1 +\sum_{i\in\mathrm{rank}(M)\setminus2} b_i= 0$implies$\lambda_0=\lambda_0+\lambda_1=0$implies$\lambda_0=\lambda_1=0$). This implies that any free$R$-module$M$with rank$\geq 2$has more than one basis. It remains to analyse the situation in which$M$is a free rank-one module over a commutative ring$R$; let any such$M$be given. Suppose that$M$has precisely one basis$\{b\}$. Then, evidently: for each$r\in R\setminus\{1\}$,$\{r\cdot b\}$is not a basis, or is equal to$\{b\}$. Hence for each$r\in R$, if$r$is a unit, then$\{r\cdot b\}$evidently is a basis, hence$r\cdot b = b$. if$r$is not a unit, then$\{r\cdot b\}$is not a basis. For further reference, we now know that • for each unit$r$of$R$we have$r\cdot b = b$. Yet after these obvious arguments, there does not seem any obvious further conclusion, and the question seems more or less reduce to if and how one can characterize commutative rings with trivial group of units. Am I missing something obvious which answers the question? • They must have characteristic$2$. – Lord Shark the Unknown Sep 23 '17 at 15:42 • @LordSharktheUnknown: yes, I know. More precisely, not "they", but the ring$R$must have characteristic 2, because otherwise$\{-b\}\cup B\setminus\{b\}\$ would be another basis. But this does not give a characterization. But thanks for pointing out. – Peter Heinig Sep 23 '17 at 15:49
• You have mentioned that the ring in question is an algebra over the field of two elements and have no more units than the field. Since there are many such rings (for example, polynomial rings over the field of two elements, but by no means the only ones), what other type of characterization are you looking for? – Mohan Sep 24 '17 at 18:21
• @Mohan: thanks for pointing out. Re "what other type of characterization are you looking for?": I don't know. Perhaps the abundance of the rings that you mention is the reason why there is nothing more to say. I just had found this an irritating 'blind spot' in a larger piece I am writing, which I could gloss over, yet am trying not to. I don't have more to say on this currently. – Peter Heinig Sep 24 '17 at 18:47