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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!

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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
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
Dear Isaac: Does $R$ have a $1$? – Pierre-Yves Gaillard Feb 23 '12 at 4:24
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
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 $1$.

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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 .

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.

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.

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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

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