7
$\begingroup$

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!

  • 0
    @Matt: In the notation I'm familiar with, I write $M = Rm$ to indicate that $m$ generates $M$, not that it is a basis.2012-02-23

2 Answers 2

8

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 $1$.

  • 0
    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.2012-02-23
3

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.

  • 0
    Dear @Isaac, your proof is correct and simple. I am not in a position to judge if it is "much simpler" or a little simpler or just as simple as mine.2012-02-23