Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a ring, and $M$ a nontrivial cyclic, free $R$-module. Let $m$ generate $M$, so that $M = Rm$. Is it then the case that $m$ forms a basis for $M$, so that $\mbox{ann}_{R}(m) = (0)$?

I know that if $R$ is a domain or a commutative ring, it is easy to show that $m$ forms a basis. However, I am unsure as to whether or not it holds for general rings. Any insight would be appreciated. Thanks!

share|improve this question
    
You accidentally the case $M=0$, $m=0$. –  darij grinberg Feb 23 '12 at 3:57
    
I'm not sure what you are suggesting. –  Isaac Solomon Feb 23 '12 at 4:04
2  
That the problem needs a modification, even if a trivial one. The module $0$ is cyclic and free (over the empty set!), but its zero does NOT form a basis (unless $R=0$). –  darij grinberg Feb 23 '12 at 4:07
3  
Dear Isaac: Does $R$ have a $1$? –  Pierre-Yves Gaillard Feb 23 '12 at 4:24
3  
It is true that a free module is a module with a basis, but it is not necessarily the case that the basis will be the generating element. A priori, the generating element might have a nontrivial annihilator, and so the necessary basis would need to have more than one element. –  Isaac Solomon Feb 23 '12 at 4:51

2 Answers 2

up vote 8 down vote accepted

Let's take Georges proof and turn it into a counterexample (!)

Let $k$ be a field and $R$ the quotient of the free algebra $k\langle x,y, z\rangle$ by the ideal generated by $xy-1$ and $zy$. As a $k$-vector space, $R$ has a basis consisting of those non-commutative monomials which contain neither $xy$ nor $zy$ as subwords —this follows immediately from Bergman's Diamond Lemma, for example, or from a simple ad hoc argument (which surely will boil down to the Diamond lemma...)

Now $M=R$, viewed as a left $R$-module as usual, is generated by $m=y$, but of course $z\cdot m=0$, so $\{m\}$ is not a basis because $m$ has a non-trivial annihilator.

Notice that $M$ is of course free of rank $1$.

share|improve this answer
    
If you like rings without $1$, consider the quotient of the non-unital $k\langle x,y,z\rangle$ by the ideal generated by $xxy-x$, $yxy-y$, $zxy-z$ and $zy$. –  Mariano Suárez-Alvarez Feb 23 '12 at 8:48
    
I hope it is clear to everybody that my proof has no counterexample if you stick to its hypothesis, made explicit now, that $R$ is commutative. –  Georges Elencwajg Feb 23 '12 at 9:08

Yes, if $m$ generates $M$, it is a basis for $M$, if $R$ is commutative .

Proof
Let $b$ be a basis of $M$, so that in particular $Ann(b)=0$.
Since $m$ generates $M$ we can write $b=rm$ for some $r\in R$.
On the other hand we can write $m=sb$ for some $s\in R$ since $b$, a basis, certainly generates $M$.
So we have $b=rm=rsb$, hence $(1-rs)b=0$ and thus $1-rs=0$ because $Ann(b)=0$.
We see that $r,s\in R^*$ are invertible and since $m=sb$ and $b$ is a basis, $m$ is a basis too.

Edit
I have used that a basis of a non-zero cyclic free module has just one element.
Since Isaac asks why in a comment, I'll give a proof.
I claim that if $g$ is a generator of $M$, any two elements on $M$ are linearly dependent (still assuming $R$ commutative !)
Indeed, if $u=ag$ and $v=bg$ are arbitrary in $M$, we have a linear relation $bu-av=0$ and either this is a nontrivial linear relation and $u,v$ are linearly dependent or $a=b=0$ and then $u=v=0$ are certainly linearly dependent in that case too.

Important new edit
I had assumed in my proof that $R$ is commutative without saying so. I have now made this assumption explicit: all my apologies to all and thanks to Mariano for calling my attention to this point.

share|improve this answer
    
Doesn't this assume that the basis for $M$ has only one element? Why can't $M$ be free on a larger basis? –  Isaac Solomon Feb 23 '12 at 8:06
    
You want to check that $sr=1$, too, no? –  Mariano Suárez-Alvarez Feb 23 '12 at 8:12
    
Dear @Isaac: yes I have assumed that because it always holds . I have given a detailed proof in an edit. –  Georges Elencwajg Feb 23 '12 at 8:34
    
You are assuming the ring is commutative in your edit (There are rings such that as left modules $R\cong R\oplus R$!) –  Mariano Suárez-Alvarez Feb 23 '12 at 8:37
    
Dear @Mariano: I had assumed, without saying so explicitly, that R is a commutative ring: professional deformation of an algebraic geometer! I have written a a new edit to clarify this . Thanks a lot for drawing my attention to my unjustified implicit assumption . –  Georges Elencwajg Feb 23 '12 at 9:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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