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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 $M=0$, $m=0$.2012-02-23
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    I'm not sure what you are suggesting.2012-02-23
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    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 $R=0$).2012-02-23
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    Oh, I see. I'll edit it accordingly. Thanks!2012-02-23
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    Dear Isaac: Does $R$ have a $1$?2012-02-23
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    @Pierre: I have not necessarily assumed so, but I would certainly be interested in a proof for unital rings.2012-02-23
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    I'm confused, Wikipedia (for instance) defines a free module as a module with basis. Maybe think about the $\mathbb{Z}$-module $\mathbb{Z}/6\mathbb{Z}$. It is generated by $\bar1$, hence cyclic, but not free as $\bar{2}=1\cdot\bar{2}=4\cdot\bar{2}$.2012-02-23
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    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.2012-02-23
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    @IsaacSolomon, ahh... so a weaker question would be if cyclic and free implies rank 1?2012-02-23
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    The case of there being a one-element basis that is not the generator is another, interesting possibility, and I see no easy way to rule it out. On the side, I'm not sure if the term "rank" is defined for rings without invariant basis number.2012-02-23
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    Here is what is confusing to me. M cyclic seems to mean that M is cyclic as a group. When you say "let m generate M" it seems implied that M = as a group. You then use "so" which seems to indicate that logically it follows that M=Rm, but writing this makes it seem like m is a basis for M as an R-module (I mean, that is standard notation for indicating a basis). I think this is some of the confusion. The wording of the question states as a fact what you are asking as a question.2012-02-23
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    @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

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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 $zy$.2012-02-23
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    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
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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.

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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?2012-02-23
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    You want to check that $sr=1$, too, no?2012-02-23
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    Dear @Isaac: yes I have assumed that because it always holds . I have given a detailed proof in an edit.2012-02-23
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    You are assuming the ring is commutative in your edit (There are rings such that as left modules $R\cong R\oplus R$!)2012-02-23
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    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 .2012-02-23
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    @Georges: If $R$ is commutative, there is a much simpler proof that the generating element forms a basis. I.e. let $b$ be any element of the basis, and $k$ a nontrivial element in the annihilator of $m$, the generating element. Write $b = rm$. Then $kb = krm = r(km) = r0 = 0$, a contradiction.2012-02-23
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    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